Zill-A First Course in Complex Analysis Some Soln2
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1.1
Aaron Morrisett
Monday, June 01, 2009 9:38 AM
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1.2 Monday, June 01, 2009 12:44 PM
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1.3 Monday, June 01, 2009 8:37 PM
Aaron Morrisett Math 459 Homework # 2
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1.4 Monday, June 01, 2009 8:37 PM
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1.5 Wednesday, June 03, 2009 9:31 AM
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2.1 Thursday, June 04, 2009 9:37 AM
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2.2 Thursday, June 04, 2009 9:37 AM
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2.3 Thursday, June 04, 2009 2:07 PM
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2.6 Monday, June 08, 2009 3:43 PM
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3.1 Tuesday, June 09, 2009 8:08 AM
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3.2 Thursday, June 11, 2009 8:05 AM
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4.3 Wednesday, June 17, 2009 9:47 AM
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5.1 Thursday, June 18, 2009 9:31 AM
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5.2 Friday, June 19, 2009 9:14 AM
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5.3 Sunday, June 21, 2009 7:15 PM
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cx
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5.4 Monday, June 22, 2009 10:45 AM
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5.5 Monday, June 22, 2009 2:07 PM
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6.2 Friday, June 26, 2009 8:08 AM
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6.3 Friday, June 26, 2009 11:42 AM
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6.4 Monday, June 29, 2009 9:39 AM
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6.5 Monday, June 29, 2009 11:55 AM
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6.6 Tuesday, June 30, 2009 9:20 AM
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Joseph Heavner Honors Complex Analysis Assignment 3 April 6, 2015
2.5 Reciprocal Function 1.) Find the image of the circle |z| = 5 under the reciprocal function mapping. This is classical inversion in a unit circle, in which the argument is invariant but the modulus is transformed to be the reciprocal of that of the domain. In other words, here our modulus becomes 1/5, making our image 1 The Circle |w| = 5
3.) Find the image of the semicircle |z| = 3 , −π/4 ≤ arg(z) ≤ 3π/4 under the reciprocal function mapping. Again, our function inverts the moduli to be 1/3, but here we too have reflection about the x-axis plays a non-trivial role. In particular, this inverts (additive) the arguments to be −3π/4 ≤ arg(w) ≤ π/4. Thus, we have The Semicircle |w| =
1 3π π , − ≤ arg(w) ≤ 3 4 4
5.) Find the image of the annulus 1/3 ≤ |z| ≤ 2 under the reciprocal function mapping. As with (1) reflection about the x-axis has no bearing on the image (fully circular regions are invariant under rotation). However, it remains that the moduli are inverted (multiplicative), thus we have The Annulus
1 ≤ |w| ≤ 3 2
7.) Find the image of the ray arg(z) = π/4 under the reciprocal function mapping. Suppose we let z = reiθ , then our domain is given by r > 0 , θ = π/4. (Note that r cannot be negative in any case, and the case where r is zero includes all arguments and so is ignored here.) If we invert r then we have 1/r > 0, but because this includes the set of all possible r, we have an invariant (intuition of the fact will here suffice, though a proof is accessible). Now, we also know that our mapping inverts arguments, in particular θ = π/4 7−→ θ 0 = −π/4. In conclusion, we have The Ray arg(w) = −
π 4
9.) Find the image of the line y = 4 under the reciprocal function mapping. This problem fits a general form that states that a line y = k, under the reciprocal function, maps to the 1 circle |w + 21 k| = | 2k |. With that in mind, the image is clearly 1 1 |w + i| = 8 8 1
10.) Find the image of the line x =
1 6
under the reciprocal function mapping.
Similar to (9) we have a problem of the form: find the image of the line x = k under the reciprocal mapping, which has been shown to be answered by
|w −
1 1 |=| | 2k 2k
Thus, we arrive at our answer
|w − 3i | = 3
11.) Find the image of the circle |z + 1| = 1 under the reciprocal mapping. As mentioned in the Remarks on page 96 and can be seen from the results regarding have lines as 1 1 i | = | 2k | to the line y = k. So, we domains, this reciprocal mapping maps a circle of the form |z + 2k identify k = 1/2, implying that we map to the line y=
1 2
15.) Find the image of the set S under the mapping w = 1/z on C ∪ {∞}
. Figure 1: The Set S We already know how to map lines to circles via the reciprocal function. So, we simply identify these as the lines x = −2 and x = −1, which map to 1 1 1 1 |w + | = ; |w + | = 4 4 2 2 We now check a point within the domain such as, say, z = −3/2, which maps to w = −2/3. Thus, we have the set containing the point w = −2/3 and bounded by the circles 1 1 1 1 |w + | = ; |w + | = 4 4 2 2
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23.) Show that the image of the line x = k, x 6= 0, under the reciprocal map defined on the extended complex plane is the circle 1 1 |w − | = | | 2k 2k The vertical line x = k consists of all points z = k + iy such that x, y ∈ R where k 6= 0. So we may write w=
1 y k − 2 i = 2 k + iy k + y2 k + y2
Thus our real and imaginary parts of w = f (u, v) are u= But observe that v =
−yu k ,
k −y , v= 2 , y∈R k 2 + y2 k + y2
which implies
vk u And so, upon substituting into our initial expression for u, we have y=−
u=
(1)
k k2
+ ( −uvk )2
It is here advantageous to us to simplify and complete the square as follows u=
k
k2 + ( −uvk )2 −vk 2 2 k = u k +( ) u v2 k 2 u k 2 u2 + v2 k 2 u u2 + v2 − k u 1 2 2 u − + − + v2 k 2k 1 2 u− + v2 2k
k = k2 u + uk =
0= 2 1 − = 2k 1 = 4k2
Finally, by recognition, we have 1 1 |w − k| = | k| 2 2 Note that we never made the restriction in (1) that u 6= 0. This is because here we are working in the extended complex plane.
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2.6 Limits and Continuity Note: In order to save time, the extreme pedantism and wordiness that is seen in the text has here been avoided. For instance, substitution to evaluate a limit may take one line rather than half a page. 1.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim (z2 − z)
z→2i
Let us simply substitute in z = 2i as follows:
(2i )2 − 2i = 4i2 + 2i = −4 + 2i
3.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim (|z|2 − i z¯ )
z →1− i
Again, let us try substitution
|1 − i |2 − i (1 − i ) = 2 − i + i 2 = 2 + 1 − i = 3 − i
5.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim (ez )
z→πi
Again, we need only substitute. eiπ = −1
9.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim (z2 − z)
z →2− i
We try substitution as follows
(2 − i )2 − 2 + i = 4 − 4i + i2 + i − 2 = 4 − 2 − 1 − 3i = 1 − 3i
11.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute 1 lim (z + ) z
z→eiπ/4
Yet again, we need only substitution for the following limit. In particular,
√ 1 lim (z + ) = eiπ/4 + e−iπ/4 = cos(π/4) + i sin(π/4) + cos(−π/4) + i sin(−π/4) = 2 iπ/4 z z→e
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13.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim
z→−i
z4 − 1 z+i
For the first time thus far, substitution will not suffice, for, obviously, it yields an indeterminate form (note that L’Hopital’s rule applies only to the reals). However, if we simply recognize that z4 − 1 = (z2 − 1)(z2 + 1) = (z − 1)(z + 1)(z + i )(z − i ), then we have that lim
z→−i
z4 − 1 = lim (z + 1)(z − 1)(z − i ) = (−i + 1)(−i − 1)(−2i ) = 4i z+i z→−i
19.) Consider the limit lim
z 2 z
z →0
a.) What value foes the limit approach as z approaches 0 along the real axis? b.) What value does the limit approach as z approaches along the imaginary axis? c.) Do the answers from (a) and (b) imply that the limit exists? Explain. d.) What value does the limit approach as z approaches along the line y = x? e.) What can you say about the limit in general? a.) Along the x axis we have z = x + 0y = x, thus the limit is lim
x →0
x 2
= lim 1 = 1
x
x →0
b.) Along the imaginary axis we have x = 0, implying that z = iy, thus our limit becomes lim
y →0
yi −yi
2
= lim (−1)2 = 1 y →0
c.) No. The limit must be the same along any of the infinitely (uncountably) many paths in the complex plane for it to exist in general. In other words, two, three, or even a trillion paths, while perhaps suggesting that the limit may exist and equal some value c, do not actually demonstrate that the limit is c. This must be shown using more general methods. However, if the limit along one path does not equal the limit along another, then we may say that the limit does not exist. d.) If we approach along y = x then z = x + ix and lim
x →0
x + xi x − xi
2
= lim
x →0
1+i 1−i
2
=
2i = −1 −2i
e.) As discussed in part (c), the limit does not exist because the limit is dependent on path.
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21.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute lim
z→∞
z2 + iz − 2 (1 + 2i )z2
Let us attempt to simplify our expression, as its current form yields an indeterminate form and there is no natural idea of the top ”approaching infinity faster” than the bottom as there is in R (this is more formally shown by dividing by the largest power of the variable in question), at least not when we have an imaginary number in question. So, upon simplification we have
(1 − 2i )z2 + (2 + i )z − (2 − 4i ) 5z2 Now that all factors are properly separated with the denominator a real expression of a complex variable, we may use our familiar laws of rational functions as apply in R. In particular, it is found that lim f (z) =
z→∞
1 2 − i 5 5
23.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute lim z →i
z2 − 1 z2 + 1
By substitution we have that, noting the nature of complex infinity (e.g. its lack of sign) z2 − 1 −2 → → ∞ 2 0 z +1
27.) Show that f is continuous at z0 f (z) = z2 − iz + 3 − 2i ; z0 = 2 − i We have that f (2 − i ) = 5 − 8i and similarly lim (z2 − iz + 3 − 2i ) = (2 − i )2 − i (2 − i ) + 3 − 2i = 4 − 4i − 1 − 2i − 1 + 3 − 2i = 5 − 8i
z →2− i
Therefore, by definition f is continuous at z0 . 28.) Show that f is continuous at z0 f ( z ) = z3 −
1 ; z0 = 3i z
Similar to (27) we know that f (z0 ) = − 80i 3 and the limit at that point is the same by virtue of substitution, i.e. 1 1 1 −80 lim f (z) = (3i )3 − = 27i3 − = −27i − = z → z0 3i 3i 3i 3i Thus f is continuous at z0 .
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29.) Show that f is continuous at z0 f (z) =
z3 ; z0 = i + 3z2 + z
z3
Substitution here works for computing the limit, and substitution is equivalent in computation to evaluating f (z0 ), thus the two are equal. Here is the explicit calculation: lim f (z) =
z → z0
i3 −i −i 1 = = = i 3 2 −i − 3 + i −3 3 i + 3( i ) + i
And so f is continuous at z0 .
31.) Show that f is continuous at z0 ( f (z) =
z3 −1 z −1
3
: |z| 6= 1 : |z| = 1
where z0 = 1 Clearly, f (1) = 3 and so we simply try the limit as follows (note the expression we chose to evaluate – this was done because it is equivalent to the value obtained along the other possible path) z3 − 1 (z − 1)(z2 + z + 1) = lim = lim (z2 + z + 1) = 3 ( z − 1) z →1 z − 1 z →1 z →1 lim
Therefore, it has been demonstrated that f is continuous at z0 = 1.
35.) Show that f is discontinuous at z0 f (z) =
z2 + 1 ; z0 = − i z+i
Evaluating f (−i ) yields
(−i )2 + 1 =∅ −i + 1 Thus, the criteria for continuity are not met and f is discontinuous at z0 .
37.) Show that f is discontinuous at z0 f (z) = Arg(z) ; z0 = −1 The problem here is not that f does not exist but that in fact the limit does not exist, thus not meeting the criteria for continuity, and so making f discontinuous at z0 . Let z be a point on the negative real axis, then Arg(z) = π but we have that there are points {zn } such that they are arbitrarily close to z but have their image under f has some nonzero imaginary part and so the argument becomes arbitrarily close to −π. Thus, we have that the limit does not exist. (An e − δ argument would make this more precise — Also, note that we must be careful with substitution in the case of poorly behaved functions like this)
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39.) Show that f is discontinuous at z0 ( f (z) =
z3 −1 z −1
3
: |z| 6= 1 : |z| = 1
where z0 = i If we consider the approach along the imaginary axis then we have lim z → i f (z) =
i3 − i =i i−1
However, trivially we know that along the unit circle we have that the limit evaluates to 3. But, 3 6= i, thus f is discontinuous at the point in question.
41.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous. z0 f (z) = Re(z) Im(z) Let z = x + iy, then f (z) = Re( x + iy) Im( x + iy) = xy But, we know that the function f (z) = xy is continuous for arbitrary z, thus f is continuous on all of C. 43.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous. z−1 f (z) = zz¯ − 4 Let z = x + iy, then f (z) =
x + iy − 1 x−1 y +i 2 = 2 ( x + iy)( x − iy) − 4 x + y2 − 4 x + y2 − 4
Therefore, u( x, y) =
x2
x−1 y ; v( x, y) = 2 2 +y −4 x + y2 − 4
These expressions are continuous on their domains. In particular, they are continuous for all ( x, y) : x2 + y2 6= 4, thus f is continuous for all z : |z| 6= 2.
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3.1 Differentiability and Analyticity 3.) Use definition 3.1 to find f 0 (z) where f (z) = iz3 − 7z2 The definition f 0 (z) = lim
∆z→0
f (z + ∆z) − f (z) ∆z
becomes i (z + ∆z)3 − 7(z + ∆z)2 − iz3 + 7z2 ∆z ∆z→0 i (z3 + 3z2 ∆z + 3z + ∆z3 ) − 7(z2 + 2z∆z + ∆z2 ) − iz3 + 7z2 = lim ∆z ∆z→0 ∆z(3iz2 + 3iz∆z + i (∆z)2 − 14z − 7∆z) = lim ∆z ∆z→0 = lim 3iz2 + 3iz∆z + i (∆z)2 − 14z − 7∆z
f 0 (z) = lim
∆z→0 2
= 3iz − 14z Thus, f 0 (z) = 3iz2 − 14z
5.) Use definition 3.1 to find f 0 (z) where f (z) = z − f 0 (z) = lim
1 z
z + ∆z −
1 z+∆z
∆z z∆z + z2 + 1 = lim ∆z→0 z ( ∆z + z ) ∆z→0
z2 + 1 z2 1 = 1+ 2 z
=
Thus, f 0 (z) = 1 +
9
1 z2
−z+
1 z
9.) Use alternate definition (12) to find f 0 (z) where f (z) = z4 − z2 The definition mentioned is restated below for completeness: f 0 (z0 ) = lim
z → z0
f ( z ) − f ( z0 ) z − z0
With that, we identify the pieces and substitute, then solve as follows on the next page: z4 − z2 − z40 + z20 z → z0 z − z0 z4 − z40 z2 − z2 = lim + 0 z → z0 z − z 0 z − z0 2 2 (z + z0 )(z − z0 )(z + z0 ) (z + z0 )(z0 − z) − = lim z → z0 z − z0 z0 − z
f 0 (z0 ) = lim
= lim z30 + z20 z + z0 z2 + z3 − z − z0 z → z0
= z30 + z20 (z0 ) + z0 (z0 )2 + (z0 )3 − (z0 ) − z0 = 4z30 − 2z0 Note, however, that this is a general point, so we may replace z0 , and so we arrive at our solution f 0 (z) = 4z3 − 2z
19.) The function f (z) = |z|2 is continuous at the origin. (a) Show that f is differentiable at the origin. (b) Show that f is not differentiable at any point z 6= 0. (a) If we take |z| to be zz then we can easily demonstrate this
(z + ∆z)(z + ∆z) − zz ∆z ∆z→0 ∆z = lim z + ∆z + z ∆z→0 ∆z
f 0 (z) = lim
At this point we may stop, noting that if z = 0, then all terms vanish and so the function is differentiable with derivative equal to zero. (b) Given the final limit of (a) we see that the term ∆z ∆z is problematic. Noting that the expression can be rewritten as an exponential. If this is done we note that the angle φ could be anything, regardless of how arbitrarily close ∆z is to zero. Thus, this portion of the limit justifies the lack of differentiability of f (z) at any point z0 ∈ C : z0 6= 0.
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21.) Show that f (z) = z is nowhere differentiable. We consider the definition of the derivative, here letting ∆z = w. lim
w →0
x − yi z+w−z w = lim = lim w w →0 w x,y→0 x + yi
This is clearly non-existent, for along the line x = 0 it evaluates to −1, whereas it is 1 along y = 0.
23.) Use L’Hopital’s rule to compute lim z →i
z7 + i z14 + 1
If we rewrite the numerator as P(z) and the denominator as Q(z) then we see that f (z) must be analytic at z = i, because it is the quotient of analytic functions (polynomials). Thus, we differentiate both P(z) and Q(z). lim z →i
i 7z6 7( i )6 −7 = = = 13 13 14i 2 14z 14(i )
27.) Determine the points at which f (z) =
iz2 − 2z is not analytic. 3z + 1 − i
We can take the derivative of the function to begin. To save paper, the work here will be somewhat ignored. Regardless, the conclusion is that f 0 (z) =
(−2 + 2i )(2 + 2i )z + 3iz2 ((1 − i ) + 3z)2
Now, we see if this is undefined anywhere, and indeed ((1 − i ) + 3z)2 = 0 at z = − 31 + 13 i, thus the function cannot be analytic there. But, because the function’s derivative exists at all other points z ∈ C we have the function is analytic everywhere except at z = − 13 + 13 i.
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3.2 Cauchy–Riemann Equations 1.) Given that f is analytic, show that the Cauchy-Riemann equations are satisfied for f (z) = z3 . By expanding and grouping we rewrite f (z) = z3 as f (z) = ( x3 − 3xy2 ) + i (3x2 y − y3 ) we take the partial derivatives: ∂u ∂v ∂v ∂u = 3x2 − 3y2 ; = −6xy ; = 6xy ; = 3x2 − 3y2 ∂x ∂y ∂x ∂y Thus the equations
∂u ∂v ∂u ∂v = and =− are satisfied. ∂x ∂y ∂y ∂x
3.) Show that f (z) = Re(z) is nowhere analytic. If we let z = x + iy then f (z) = x, thus u( x, y) = x and v( x, y) = 0. Therefore, ∂u ∂u ∂v ∂v =1; =0; =0; =0 ∂x ∂y ∂x ∂y Because the Cauchy-Riemann equations are not satisfied (in particular ∂u/∂x 6= ∂v/∂y) for no points in the complex plane. So, we may conclude that f (z) is nowhere analytic. 7.) Show that f (z) = x2 + y2 is nowhere analytic. u( x, y) = x2 + y2 and v( x, y) = 0. Computing the partials of f : ∂u ∂u ∂v ∂v = 2x ; = 2y ; =0; =0 ∂x ∂y ∂x ∂y Therefore the Cauchy-Riemann equations are only satisfied at the point z = 0 in the complex plane. However, there does not exist a domain R with z = 0 in R in which there all points in R satisfy the Cauchy-Riemann equations. Therefore, f is nowhere analytic.
8.) Show that f (z) = u( x, y) =
x x 2 + y2
x y +i 2 is nowhere analytic. x 2 + y2 x + y2
and v( x, y) =
y , x 2 + y2
which means that
∂u y2 − x 2 ∂u −2xy ∂v −2yx ∂v x 2 − y2 = 2 ; = ; = ; = ∂x ( x + y2 )2 ∂y ( x2 + y2 )2 ∂x ( x2 + y2 )2 ∂y ( x 2 + y2 )2 We observe that ∂u/∂x = ∂v/∂y if and only if y = ± x where x 6= 0. However, ∂u/∂y = −∂v/∂x if and only if x = 0 where y 6= 0 or y = 0 where x 6= 0. Clearly, these two conditions are mutually exclusive, ensuring that the Cauchy-Riemann equations are not satisfied, implying that f is nowhere analytic.
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9.) (a) Use Theorem 3.2.2 to show that f (z) = e− x cos y − ie− x sin y is analytic in an appropriate domain, and (b) find the derivative of f in said domain using (9) or (11). (a) u( x, y) = e− x cos y and v( x, y) = −e− x sin y, therefore ∂u ∂u ∂v ∂v = −e− x cos y ; = −e− x sin y ; = e− x sin y ; = −e− x cos y ∂x ∂y ∂x ∂y Clearly the Cauchy-Riemann equations are satisfied for all z ∈ C, making f analytic in all of the complex plane. (b) It is here advantageous to use ”Cartesian” (perhaps the term ”Argand” or rectangular is better here) coordinates, so we compute f 0 (z) by using (9), which states that f 0 (z) =
∂v ∂v ∂u ∂u +i = −i ∂x ∂x ∂y ∂y
Via substitution (note that here the book appears to have the incorrect answer) f 0 (z) = −e− x cos y + ie− x sin y
2
2
2
2
11.) (a) Use Theorem 3.2.2 to show that f (z) = e x −y cos(2xy) + ie x −y sin(2xy) is analytic in an appropriate domain, and (b) find the derivative of f in said domain using (9) or (11). (a) u( x, y) = e x
2 − y2
cos(2xy) and v( x, y) = 2x
2 − y2
sin(2xy), so
2 2 2 2 ∂u ∂u = 2e x −y ( x cos(2xy) − y sin(2xy)) ; = −2e x −y (y cos(2xy) + x sin(2xy)) ∂x ∂y 2 2 2 2 ∂v ∂v = 2e x −y (y cos(2xy) + x sin(2xy)) ; = 2e x −y ( x cos(2xy) − y sin(2xy)) ∂x ∂y
It is now clear that f is an entire function, i.e. it is analytic throughout all of the complex plane. (b) By formula (9) f 0 (z) = f (z) and in particular f 0 (z) = e x cos(y) + ie x sin(y)
17.) Find real constants a, b such that f (z) = 3x − y + 5 + i ( ax + by − 3) is analytic. u( x, y) = 3x − y + 5 and v( x, y) = ax + by − 3, so ∂u ∂u ∂v ∂v =3; = −1 ; =a; =b ∂x ∂y ∂x ∂y It is therefore clear that a = 1, b = 3
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19.) Show that f (z) = x2 + y2 + 2ixy is not analytic at any point but is differentiable along the x-axis, and (b) use (9) or (11) to compute the derivative along the axis. (a) u( x, y) = x2 + y2 and v( x, y) = 2xy, thus ∂u ∂u ∂v ∂v = 2x ; = 2y ; = 2y ; = 2x ∂x ∂y ∂x ∂y The first Cauchy-Riemann equation is universally satisfied, whereas the second is satisfied if and only if y = 0. Thus, while f is differentiable on the x-axis, it is not analytic, for there exists no neighborhood in which the Cauchy-Riemann equations are satisfied. (b) Via (9), f 0 (z) = 2x + i2y, which, along the x-axis where y = 0 means that f 0 (z) = 2x
21.) (a) Show that f (z) = x3 + 3xy2 − x + i (y3 + 3x2 y − y) is not analytic at any point but is differentiable along the coordinate axes, and (b) use (9) or (11) to compute the derivative along the axes. (a) u( x, y) = x3 + 3xy2 − x and v( x, y) = y3 + 3x2 y − y, meaning that ∂u ∂u ∂v ∂v = 3x2 + 3y2 − 1 ; = 6xy ; = 6xy ; = 3y2 + 3x2 − 1 ∂x ∂y ∂x ∂y The first Cauchy-Riemann equation is satisfied for all z ∈ C, whereas the second is satisfied only if x = 0, y = 0, or x = y = 0. Similar to (19), there exists no neighborhood in which the Cauchy-Riemann equations are satisfied, therefore f is not analytic, but the Cauchy-Riemann equations are satisfied on the coordinate axes, making them differentiable there. (b) In particular, f 0 (z) = 3x2 − 1 along the x −axis f 0 (z) = 3y2 − 1 along the y−axis
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Joseph Heavner Honors Complex Analysis Assignment 1 December 26, 2014
1.1 Complex Numbers and Their Properties 2.) Write the given number in the form a + bi. (a)
2i3
− 3i2
+ 5i
(b)
3i5
− i4
+ 7i3
− 10i2
−9
5 2 20 (c) + 3 − 18 i i i
6
(d) 2i +
2 −i
3
+ 5i−5 − 12i
(a) Given the identity i2 = −1 we have the following re-expression 2(−i ) − 3(−1) + 5i, which simplifies further to 3 + 3i (b) Because the powers of i are periodic with a cycle of 4 we can determine the value of in for any n ∈ Z+ by simply taking i n mod 4 . Using this, we have the following simplification 3(i ) − (1) + 7(−i ) − 10(−1) − 9 which leads us to our solution 0 − 4i (c) Let us rewrite this as
2i 5 2 20 5i + − = 2+ + 20 = −5i + 2i + 20 = 20 − 3i i (−i ) −1 −i · i i
(d) Again, we begin by rewriting 3 our exponents, but this time we too distribute exponentiation over a quotient as follows 2(−1) + (−2i)3 + 5( 1i ) − 12i = −2 + 8i + 5i − 12i = −2 − 8i − 5i − 12i = −2 − 25i i2 i2 23.) Use the binomial theorem to write (−2 + 2i )5 in a + bi form. Recall the binomial theorem. Binomial Theorem: Let x, y ∈ C, then n
n n−k k ∀n ∈ Z : ( x + y) = ∑ x y k k =0 n
+
In our situation, this is 5
5 (−2 + 2i ) = ∑ (−2)5−k (2i )k k k =0 5
Expanding this out yields the following 1 · (−2)5 (2i )0 + 5 · (−2)4 (2i )1 + 10 · (−2)3 (2i )2 + 10 · (−2)2 (2i )3 + 5 · (−2)1 (2i )4 + 1 · (−2)0 (2i )5 Upon simplication (term by term, keeping symmetry in mind) we have −32 + 160i − 320i2 + 320i3 − 160i4 + 32i5 , which, using the definition of i gives −32 + 160i + 320 − 320i − 160 + 32i, which yields our answer 128 − 128i
1
√ 35.) Show that z1 = −
√ 2 2 + i satisfies the equation z2 + i = 0. Find an additional solution, z2 . 2 2
To verify the suggested solution is indeed valid we need only substitute and finish with an identity, thus
√ !2 2 2 + i +i = 0 − 2 2 √ !2 √ √ ! √ !2 2 2 2 2 = −i +2 − · i + i − 2 2 2 2 √
1 i2 + 2(−.5i ) + = −i 2 2 1 1 − − i = −i 2 2 −i = −i Now, to find an additional solution we use a clever trick. Because the squaring function transforms all negatives into positives and because of the nature of i, we know that the solution set here will be x = { a + bi , − a − bi }
√ 2 2 − i So, we must show, like before, that we can satisfy our equation, but this time with z2 = 2 2 √ !2 √ 2 2 − i +i = 0 2 2 √ !2 √ √ ! √ !2 2 2 2 2 +2 − · i + − i = −i 2 2 2 2 √
i2 1 + 2(−.5i ) + = −i 2 2 1 1 − − i = −i 2 2 −i = −i Given that, we have our answer via recognition and, finally, direct verification:
√ z2 =
√ 2 2 − i 2 2
46.) Think of an alternative solution to Problem 24. Then without doing any significant work, evaluate (1 + i )5404 . For problem 24 an elegant solution would be as follows: (i + 1)2 = 1 + 2i − 1 = 2i so 4 (i + 1)8 = (i + 1)2 = (2i )4 = 24 · i4 = 16 Similarly, for the more daunting problem at hand we have 2702 (1 + i )5404 = (1 + i )2 = (2i )2702 = (i2 )1351 · 22702 = −2 2702 2
√ 49.) Assume for the moment that 1 + i makes sense in the complex number system. Demonstrate the validity of the equality r r √ 1 1√ 1 1√ + 1+i = 2+i − + 2 2 2 2 2 We may not have reached roots in our text yet, but we can still use exponents and eliminate the root, leaving us with something that our text has discussed. In particular, here is the verification, ignoring the negative root for now √
!2 r 1 1√ 1 1√ 1+i = + 2+i − + 2 2 2 2 2 !2 ! r r 1 1√ 1 1√ i+1 = i − + + 2 +2 2 2 2 2 2 1 1 1 1 1 i+1 = + √ +2 i + −√ 2 2 2 2 2 2
r
r
1 1√ + 2 2 2
!!
r
+ i
1 1√ − + 2 2 2
!2
1+i = 1+i Because we have reached an identity and our steps are reversible, we have successfully shown that the equality holds. Typically, one would ”clean up” this proof by starting from first principles and working backwards, but this is more transparent with respect to our methodology.
3
1.2 Complex Plane 3.) Graph z1 = 5 + 4i, z2 = −3i, 3z1 + 5z2 , and z1 − 2z2 as vectors. Here I will manually draw these rather than use software, though the latter choice would look nicer. First, however, we will find explicit expressions for the third and fourth graphs. In particular 3(5 + 4i ) + 5(−3i ) = 15 − 3i and (5 + 4i ) − 2(−3i ) = 5 + 10i.
6.) (a) Plot the points z1 = −2 − 8i, z2 = 3i, and z3 = −6 − 5i. (b) The points in (a) determine a triangle with vertices z1 , z2 , z3 . Express each side of the triangle as a difference of vectors. a.) Again, we choose to manually plot the points
b.) The side connecting points z1 and z2 is z2 − z3 , the side connecting z1 and z3 is z3 − z1 , and the side connecting z2 and z3 is z2 − z1 . This can be best seen geometrically and can be expressed as follows
− → −−→ −−→ z− 1 z2 = z2 − z1 , z1 z3 = z3 − z1 , z2 z3 = z3 − z2
4
7.) In problem 6, determine whether the points z1 , z2 , z3 are the vertices of a right triangle. In order for a triangle to be a right triangle it must obey the Pythagorean theorem a2 + b2 = c2 where a, b, c are the lengths of the three sides of the triangle. So, we must determine the distance from z1 to z2 , from z1 to z3 , and from z2 to z3 . This goes as follows q √ d1 = |z2 − z1 | = (0 − (−2))2 + (3 − (−8))2 = 125 and d2 = | z3 − z1 | =
q
(−6 − (−2))2 + (−5 − (−8))2 =
√
25 = 5
and d3 = | z3 − z2 | =
q
(−6 − (0))2 + (−5 − 3)2 =
Now, finally we have that 52 + 102 =
√
125
√
100 = 10
2
So, we have verified that the vertices form a right triangle.
11.) Find the modulus of z=
2i 3 − 4i
We have to convert this to a + bi form as follows 2i 3 + 4i 8 6i · =− + 3 − 4i 3 + 4i 25 25 Thus, we can take
|z| =
q
(−8/25)2 + (6/25)2 =
q
(64/625) + (36/625) =
√
4/25 =
2 5
17.) Describe the set of points z in the complex plane that satisfy Re((1 + i )z − 1) = 0 For z = a + bi we have Re((1 + i )( a + bi ) − 1) = Re( a + bi + ai − b − 1) = a + b − 1 Thus, we have the line a + b − 1 = 0 or, equivalently, x − y = 1, which can be rewritten in the familiar notation y = x − 1.
5
19.) Describe the set of points z in the complex plane that satisfy
| z − i | = | z − 1| Rewriting this by definition for z = x + iy we get q q x 2 + ( y − 1)2 = ( x − 1)2 + y2 x2 + y2 − 2y + 1 = x2 − 2x + 1 + y2
−2y = −2x y=x Thus, this set of points describes the line in the complex plane y=x
23.) Describe the set of points z in the complex plane that satisfy
| z − 1| = 1 This can be explained a few ways. One would be to set this up in terms of trigonometric functions and make the connection between the parametric equations governing the unit circle. Another would be to geometrically think of this in terms of the locus of all points 1 away from the center, which is at 1. More analytically, however, we can go through definitions until we get to the standard definition of the circle in the Cartesian plane with radius r and center (h, k)
( x − h )2 + ( y − k )2 = r 2 So, we use our definitions and algebra as follows to prove this is indeed what it is. For z = x + iy we have q ( x − 1)2 + y2 = 1
( x − 1)2 + y2 = 1 This describes a circle with radius r = 1 and center (1, 0). 27.) Establish the simultaneous inequality If |z| = 2, then 8 ≤ |z + 6 + 8i | ≤ 12 Let z = x + iy, and we note that |z| = 2 =⇒ x2 + y2 = 4. Now, we split our modulus in two, with |z| = 2, thus 8 ≤ 2 + |6 + 8i | ≤ 12 =⇒ 8 ≤ 12 ≤ 12 By the triangle inequality, this must be the maximum (the least upper-bound AKA limit superior) of the inequality. Similarly, we have that |z1 | − |z2 | ≤ |z1 + z2 |, thus the following expression yields the minimum (greatest lower bound or limit inferior) 8 ≤ |6 + 8i | − 2 ≤ 12 =⇒ 8 ≤ 8 ≤ 12
It has been suggested that perhaps it would be better form to combine these inequalities into one expression.
6
37.) Under what circumstance does |z1 + z2 | = |z1 | + |z2 |? Let z1 = a + bi and z2 = x + iy then we have to find under which conditions q q p ( a + x )2 + ( b + y )2 = a2 + b2 + x 2 + y2 So, we first square both sides as follows
( a + x ) + (b + y) = 2
p
a2
+ b2
q
x 2 + y2 + a2 + b2 + x 2 + y2
a2 + 2ax + x2 + b2 + 2yb + y2 = 2
p
a2 + b2
q
x 2 + y2 + a2 + b2 + x 2 + y2
2
2
q p 2( ax + yb) = 2 a2 + b2 x2 + y2 q p ax + yb = a2 + b2 x2 + y2 Re(z1 )Re(z2 ) + Im(z1 )Im(z2 ) = |z1 ||z2 | = |z1 z2 | Thus, |z1 + z2 | = |z1 | + |z2 | if and only if Re(z1 )Re(z2 ) + Im(z1 )Im(z2 ) = |z1 z2 | Remark: A concrete example of this would be if |z1 | = 0 and/or |z2 | = 0. Challenge: Think of this geometrically with vectors. 38.) Using the complex variable z, find an equation of a circle in the complex plane where θ is measured in radians from the positive x-axis. The following equation defines a circle of radius r centered at some point z0 in the complex plane
| z − z0 | = r This comes directly from the definition of a circle being the set of all points some distance d away from some center point. More precisely, as in (23), we can go through definitions to get the familiar formula for a circle, thus proving our proposition q ( x − a )2 + ( y − b )2 = r
( x − a )2 + ( y − b )2 = r 2 Which is our normal formula for a circle centered at ( a, b) with radius r. Note that we define our complex variable as z = x + iy and our complex number as z0 = a + bi. 39.) Describe the set of points z in the complex plane that satisfy z = cos θ + i sin θ, where θ is measured in radians from the positive x-axis. By Euler’s identity we have that eiθ = cos θ + i sin θ, so we could say that it would be the set of all points that can be produced from that. However, more specifically, we have that the this set of points can take on any complex number with modulus one by virtue of our representation of complex numbers as polar forms (exponential or trigonometric/circular). In other words, the set of all points satisfying this equality forms the unit circle in the complex plane. This same figure and variants of it have already been featured multiple times in this assignment, but here we provide a quick algebraic verification of the proposition above. Because this is a parameterization such that x = cos θ and y = sin θ we have that the set z of all numbers for x, y ∈ R such that z = { x + iy}. In other words, any complex number, with the exception, which we hadn’t before mentioned, that x, y ≤ 1 (a fact obtained a priori by parametric equations and the images of sine and cosine) is represented by this. 7
49.) In this problem we will start you out in the proof of the first property |z1 z2 | = |z1 ||z2 | in (3). By the first result in (2) we can write |z1 z2 |2 = (z1 z2 ) (z1 z2 ). Now use the first property in (2) of section 1.1 to continue the proof. The property aforementioned is that z1 z2 = z¯1 z¯2 So, we proceed as follows
| z1 z2 |2 = ( z1 z2 ) ( z1 z2 ) = (z1 z2 ) (z¯1 z¯2 ) = | z1 |2 | z2 |2 ¯ Now, we may root both sides of our equation to Where the last step is obtained by the fact that |z|2 = zz. obtain the desired identity, noting that |z1 | , |z2 | ∈ R+ . q q | z1 z2 |2 = | z1 |2 | z2 |2
| z1 z2 | = | z1 | | z2 |
50.) In this problem we guide you through an analytical proof of the triangle inequality (6). Since |z1 + z2 | and |z1 | + |z2 | are positive real numbers, we have |z1 + z2 | ≤ |z1 | + |z2 | if and only if |z1 + z2 |2 ≤ (|z1 | + |z2 |)2 . (a) Explain why |z1 + z2 |2 = |z1 |2 + 2Re(z1 z2 ) + |z2 |2 . (b) Explain why (|z1 | + |z2 |)2 = |z1 |2 + 2|z1 z2 | + |z2 |2 (c) Use parts (a) and (b) along with the results in Problem 46 to derive (6). a.) This derivation is pretty straightforward for z1 = x + iy and z2 = a + bi. Below is the work.
( x + a)2 + (y + b)2 = x2 + 2ax + a2 + y2 + 2by + b2 = a2 + b2 + 2( ax + by) + x2 + y2 = |z1 |2 + 2Re(z1 z2 ) + |z2 |2 b.) Similarly, we get this result by mere algebraic manipulation without much thought. q
x2
+ y2
+
p
a2
+ b2
2
=2
p
a2 + b2
q
x 2 + y2 + a2 + b2 + x 2 + y2 = | z1 |2 + 2| z1 z2 | + | z2 |2
c.) Given those results and the fact that |Im(z)| ≤ |z| and |Re(z)| ≤ |z| we can prove the triangle inequality, which geometrically states that the length of two sides of a triangle must be greater than the remaining side. Taking (a) and (b) we see that, because 2Re(z1 z2 ) ≤ 2z1 z2 by the result in Problem 46,
|z1 + z2 |2 ≤ (|z1 | + |z2 |)2 Thus, it easily follows that, upon square-rooting both sides,
| z1 + z2 | ≤ | z1 | + | z2 | 8
1.3 Polar Form of Complex Numbers 1.) Write z = 2 in polar form, first using an argument θ 6= Arg(z) and then using θ = Arg(z). p First, we find the modulus |2| = (2)2 + (0)2 = 2 and the principle argument θ = arctan(0/2) = 0. Now, using our polar form we have 2cis(2π ) and 2cis(0) Note our introduction of the ”cis” notation for polar form. The convenience of this notation makes it optimal, especially while we have not yet been introduced formally to the exponential form of a complex number. However, notations will be varied throughout for no reason other than diversity. 3.) Write z = −3i in polar form, first using an argument θ 6= Arg(z) and then using θ = Arg(z). p Our modulus is (0)2 + (3)2 = 3 and our argument is arctan −03 = − π2 + 2πk , k ∈ Z. Thus, the following two are polar forms of z, with the first using an arbitrary argument (co-terminal to the principle argument) and the second being the principle argument. ( 3 cos − π2 + i sin − π2 z= 3π 3 cos 3π 2 + i sin 2
√ 7.) Write z = − 3 + i in polar form, first using an argument θ 6= Arg(z) and then using θ = Arg(z). q √ √ Our modulus is |z| = (− 3)2 + (1)2 = 2 and our argument is arctan(−1/ 3) = 5π/6 + 2πk, so we have our two answers ( z=
9.) Write z =
3 −1+ i
2 cis(−7π/6) 2 cis(5π/6)
in polar form, first using an argument θ 6= Arg(z) and then using θ = Arg(z).
First we need z in a + bi form, so we begin by multiplying by the conjugate on both top and bottom then we simplify −3 − 3i 3 3i =− − 2 2 2 √ p − 23 3 2 − 1 2 2 Now we find the modulus |z| = (−3/2) + (−3/2) = 2 and principle argument tan = 3 −2
arctan (1) =
− 3π 4 .
Thus, we have our two answers ( z=
√ 3 2 2 √ 3 2 2
3π cos − 3π 4 + i sin − 4 5π cos 5π 4 + i sin 4
9
√ √ 11.) Use a calculator to write z = − 2 + 7i in polar form, first using an argument θ 6= Arg(z) and then using θ = Arg(z). q √ √ As before we find the modulus to be ( 7)2 + (− 2)2 = 3 and, with the aid of a calculator, we find our √ √ argument to be arctan(− 7/ 2) ≈ 2.06168, making our solution 3 cis(8.34486) and 3 cis(2.06168)
13.) Write the complex number z whose coordinates (r, θ ) are (4, −5π/3) in the form a + bi. In √ other words, z = 4 cis(−5π/3), which is the same as 4 cos(−5π/3) + 4i sin(−5π/3) = 4(1/2) + 4i ( 3/2). So, our final answer is √ 2 + 2 3i 15.) Write the complex number 7π 7π z = 5 cos + i sin 6 6 in the form a + bi. Our trigonometric√form is already partitioned into two parts, a and b, upon distribution, thus we have 5 3 7π 5 a = 5 cos 7π 6 = − 2 and b = 5 sin 6 = − 2 , so we have
√ 5 3 5 z=− − i 2 2
19.) Use (6) and (7) to find z1 z2 and z1 /z2 in the form a + bi where π π 3π 3π z1 = 2 cos + i sin , z2 = 4 cos + i sin 8 8 8 8 This requires few words, so here are the computation via (6) and (7): z1 z2 = 8 cis(π/2) = 8 cos(π/2) + 8i sin(π/2) = 8i and
√ √ √ √ z1 1 1 2 1 2 2 2 = cis(−π/4) = .5 cos(−π/4) + .5i sin(−π/4) = + i = − i z2 2 2 2 2 2 4 4
10
21.) Write z in polar form. Then use either (6) or (7) to obtain the polar form of the given number. Finally, write the polar form in the form a + bi. √ z = (3 − 3i ) 5 + 5 3i I feel as though this question may be incorrectly written, or at least obscurely phrased. Anyway, based on the answer key in the back of the text, the question is fairly forward with the exception that we split the complex number z into its divisors z1 and z2 . p √ For z1qwe have modulus 32 + (−3)2 = 18 and argument arctan(−1) = 7π/4. For z2 we have mod√ √ √ ulus 52 + (5 3)2 = 10 and argument arctan( 3) = π/3. So, we have z1 = 18 cis(7π/4) and z2 = 10 cis(π/3), meaning that our whole number √ z = z1 z2 = 10 18 cis(π/12) In order to go back to a + bi form we simply distribute r and evaluate our trigonometric functions
√ √ √ √ 10 18 cos(π/12) + 10 18i sin(π/12) = 15( 3 + 1) + 15i ( 3 − 1)
23.) Write z in polar form. Then use either (6) or (7) to obtain the polar form of the given number. Finally, write the polar form in the form a + bi. z=
−i 1+i
p 2 2 Similar to before, we let z = z1 /z2 , where z1 = −i and z2 = 1 + i. For z1√the modulus √is 0 + (−1) = 1 and argument is (from pure geometry) 3π/2, and for z2 the modulus is 12 + 12 = 2 and the argument is arctan(1) = π/4. Thus, √ cis(3π/2) 2 z= √ cis(5π/4) = 2 2 cis(π/4) To get to a + bi form we do as follows
√
√ 1 2 2 i cos(5π/4) + i sin(5π/4) = − − 2 2 2 2
25.) Use (9) to compute
1+
√ 9 3i
√ If we take the modulus and argument of our z = 1 + 3i to convert it into trigonometric form, then we can determine the ninth power of the number via de Moivre’s formula. The modulus is given by q easily √ √ 2 (1) + ( 3)2 = 2 and the argument arctan( 3) = π/3, so, in polar form, z = 2 cis(π/3). Given z, finding z9 is simply applying a formula as follows z9 = (2 cis(π/3))9 = 29 cis(3π ) = 512(−1) = −512
11
31.) Write z in polar form and then in a + bi form. π π π 12 h π i5 2 cos + sin z = cos + i sin 9 9 6 6 Using de Moivre’s formula we get z = (cos(12π/9) + i sin(12π/9)) [32 (cos(5π/6) + i sin(5π/6))] Now, multiplying our polar forms we have z = 32 cis(13π/6) Then, distributing r and evaluating our sine and cosine functions, we have
√ √ z = 32 cos(13π/6) + 32i sin(13π/6) = 32( 3/2) + 32i (1/2) = 16 3 + 16i
33.) Use de Moivre’s formula (10) with n = 2 to find the trigonometric identities for cos 2θ and sin 2θ. This classic exercise is fairly straightforward, we simply take de Moivre’s formula for n = 2 as follows
(cos ( x ) + i sin ( x ))2 = cos (2x ) + i sin (2x ) Now we expand the binomial on the left hand side to get cos2 ( x ) − sin2 ( x ) + 2i sin (2x ) cos (2x ) = cos (2x ) + i sin (2x ) Now, we simply take Re(z) on each side to get our expression for cosine and we take Im(z) on both sides to get the expression for sine. In particular, we find that cos(2x ) = cos2 ( x ) − sin2 ( x ) and sin(2x ) = 2 sin( x ) cos( x )
12
35.) Find the positive integer n for which the following holds !n √ 3 1 = −1 + i 2 2 Because Chapter 1.4 is about roots, we unfortunately cannot simply take the roots of −1 and check. Instead, we expand out binomials, first by brute force, then by the binomial theorem. This is still fairly tedious, but it will do, especially because the n needed here is quite small. For n = 1 we have just the result inside the √ parentheses, for n = 2 we double-distribute, yielding i 2 3 . For greater n we choose to use the binomial theorem instead. For n = 3 we have √ !3− k k 3 3 3 1 i =i ∑ k 2 2 k =0 Similarly, for n = 4 and n = 5 we have, respectively, 4
4 ∑ k k =0
√ !4− k k √ 3 1 1 i 3 i =− + 2 2 2 2 and
5
5 ∑ k k =0
√ √ !5− k k 3 3 i 1 i =− + 2 2 2 2
And finally, we have that for n = 6 the equality is satisfied, because √ !6− k k 6 6 3 1 i = −1 ∑ k 2 2 k =0 Thus, for n = 6 the equality holds. (This could have been done alternatively with de Moivre’s theorem.) 39.) Suppose that z1 = r (cos θ + i sin θ ). Describe geometrically the effect of multiplying z1 by a complex number of the form z2 = cos α + i sin α, when α > 0 and when α < 0. Because the second complex number has modulus one, multiplying the moduli simply yields r, so the vector used to describe z1 z2 is equal in length to that of z1 (i.e. no scaling occurs). On the other hand, in general, we have that θ 6= α, which means that our angle changes (indeed, θ 0 = θ + α, where the prime notation is reserved, not for differentiation, but rather for the argument of z1 z2 ). For any alpha we clearly have a rotation due to the addition of arguments. But, for positive alpha we have a larger angle measured from the positive x = Re(z) axis; in other words, for positive alpha we have a counterclockwise rotation of z1 by α. Similarly, for negative alpha we have a clockwise rotation of z1 by a factor of α. 42.) Are there any special cases in which Arg(z1 z2 ) = Arg(z1 )Arg(z2 )? Prove your assertions. We know that Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ), therefore the only way we can have this is if Arg(z1 ) + Arg(z2 ) = Arg(z1 )Arg(z2 ). For z1 = x + iy and z2 = a + bi we have (y/x ) + (b/a) = (y/x )(b/a). If we let X = y/x and A = b/a then we have A + X = AX, which has solutions A, X = 2 , A, X = 0. So, in other words, if y, b = 0, i.e. the numbers are real, then the equality holds. Similarly, if y/x , b/a = 2, then the equality holds. The first case seems to have some important significance, the second seems less notable. Challenge: Think in terms of angle magnitudes and try to generalize.
13
1.4 Powers and Roots 1.) Use (4) to compute all roots of z. Give the principal nth root in each case. Sketch the roots w0 , w1 , · · · , wn−1 on an appropriate circle centered at the origin. z = (8)1/3 In this case the principal nth root is simple (the one learned in all elementary algebra classes) and evaluates to 2 by the fact that 23 = 8. For the other roots we need more powerful techniques found in our text. In particular, we begin by finding 8 in polar form. The modulus is 8 and argument 2π, from pure geometry. Therefore, given our formula for roots for n = 3, we find
(8)1/3 = 2 cis(0) , 2 cis(2π/3) , 2 cis(4π/3) = 2 , −1 ±
√
3i
9.) Use (4) to compute all roots of z. Give the principal nth root in each case. Sketch the roots w0 , w1 , · · · , wn−1 on an appropriate circle centered at the origin.
−1 +
√ 1/2 3i
q √ √ First, we write z in polar form. The modulus is (−1)2 + ( 3)2 = 2 and argument arctan(− 3) = 2π/3. So, from our root formula for n = 2 we have the following (with the first being the principal root):
√
z=
√
2 cis(π/3) ,
√
√ 2 cis(−2π/3) =
√ √ √ 2 6 2 6 + , − − ; ≈ .7071 + 1.2247i , −.7071 − 1.2247i 2 2i 2 2i
14
15.) (a) Verify that (4 + 3i )2 = 7 + 24i. (b) Use part (a) to find the two values of (7 + 24i )1/2 . a.) We expand the binomial as follows, noting the definition of the imaginary unit
(4 + 3i )2 = 16 + 24i + 9i2 = 16 − 9 + 24i = 7 + 24i
b.) By virtue of the square-root function being inverse to the squaring function, with the caveat that there are two square-roots of a number (not the principle root), one positive and one negative. Thus, (7 + 24i )1/2 is not only 4 + 3i but also −(4 + 3i ), making our simplified solution set
{4 + 3i , −4 − 3i }
17.) Find all solutions of the equation z4 + 1 = 0 By algebraic manipulation we have that the problem is to find the fourth roots of -1, which is fairly trivial when given our formula for roots √ √ θ + 2kπ θ + 2kπ n n z = r cos + i sin n n Given that, we must only find the modulus and argument of z = −1 to proceed. Arg(−1) = tan−1 p and | − 1| = (−1)2 + (0)2 = 1, thus √ √ π + 2kπ π + 2kπ 4 4 −1 = 1 cos + i sin 4 4
0 −1
=π
which yields cos(π/4) + i sin(π/4) , cos(3π/4) + i sin(3π/4) , cos(5π/4) + i sin(5π/4) , cos(7π/4) + i sin(7π/4) and simplifies to our final answer
√ 4
√
√ 2 2 −1 = ± ±i 2 2
15
19.) (a) Show that the n nth roots of unity are given by
(1)1/n = cos
2πk 2πk + i sin , k = 0, 1, 2, · · · , n − 1 n n
(b) Find the n nth roots of unity for n = 3, 4, 5. (c) Carefully plot the roots of unity found in (b). Sketch the regular polygons formed with the roots as vertices. [Hint: See (ii) in the Remarks.] a.) This is trivial by our roots p formula; we must only find the argument and modulus of z = 1 as follows Arg(1) = 01 = 0 and |1| = (0)2 + (1)2 = 1. Thus, plugging in θ = 0 and r = 1 into our formula we get
√ n
1 = cos
2πk 2πk + i sin n n
for k = 0, 1, · · · , n − 1. b.) For n = 3 we have, by substitution: cis (0) , cis
2π 3
, cis
4π 3
√ √ 1 3 1 3 = 1, − + ,− − 2 2 2 2
Similarly, for n = 4 we get: cis(0) , cis(π/2) , cis(π ) , cis(3π/2) = 1 , i , −1 , −i And, finally for n = 5 we have: cis(0) , cis(2π/5) , cis(4π/5) , cis(6/pi/5) , cis(8π/5)
= 1 , 0.30902 + 0.95106i , −0.80902 + 0.58779i , −0.80902 − 0.58779i , 0.30902 − 0.95106i c.)
16
22.) Consider the equation (z + 2)n + zn = 0, where n is a positive integer. By any means, solve the equation for z when n = 1. When n = 2. For n = 1 this simplifies to the linear equation 2z + 2 = 0 =⇒ z = −1 For n = 2 we have
( z + 2)2 + z2 = 0 z2 + 4z + 4 + z2 = 0 2z2 + 4z + 4 = 0 2(z2 + 2z + 2) = 0 Now, by the quadratic formula, which, derived from completing the square, states that for any second degree polynomial ax2 + bx + c = 0 the following holds true √ −b ± b2 − 4ac x= 2a we have the following z=
−2 ±
√ −2 ± i 4 22 − 4(1)(2) = = −1 ± i 2(1) 2
p
17
23.) Consider the equation in (22). (a) In the complex plane, determine the location of all solutions z when n = 5 [Hint: Write the equation in the form [(z + 2)/(−z)]5 = 1 and use part (a) of Problem 19.] (b) Reexamine the solutions of the equation in Problem 22 for n = 1, 2. Form a conjecture as to the location of all solutions of (z + 2)n + zn = 0. a.) As suggested, we rewrite the equation as
z+2 −z
5
= 1 =⇒ 11/5 =
z+2 −z
We know from (19) that the fifth roots of unity are
(1)1/5 = cis(0) , cis(2π/5) , cis(4π/5) , cis(6π/5) , cis(8π/5) So, we have five equations to solve, most of which involve trigonometric functions whose outputs are, well, messy. They are as follows. z+2 = 1 =⇒ z = −1 −z & q √ z+2 ≈ .30902 + .95106i =⇒ z = −1 + 5 − 2 5 ≈ −1 + 0.726541i −z & q √ z+2 ≈ .30902 − .95106i =⇒ z = −1 − 5 − 2 5 ≈ −1 − 0.726541i −z & q √ z+2 ≈ −.80902 − .58779i =⇒ z = −1 − 5 + 2 5 ≈ −1 − 3.07767i −z & q √ z+2 ≈ −.80902 + .58779i =⇒ z = −1 + 5 + 2 5 ≈ −1 + 3.07767i −z b.) Thus far all solutions have been on the line Re(z) = x = −1, i.e. all roots ω0 , ω1 , · · · , ωn have real part −1. So, we conjecture the following: Conjecture: All solutions to (z + 2)n + zn = 0 lie on the line x = −1, where Re(z) = x.
18
27.) The vector given in Figure 1.4.3 represents one value of z1/n . Using only the figure and trigonometry — that is, do not use formula (4) – find the remaining values of z1/n when n = 3. Repeat for n = 4 and n = 5. The figure shows one root to be ω0 = − 2 Thus all of our other roots will be the same distance from the origin (i.e. lie on the same circle), but rotated by 2π/3. Thus, ω1 is the number with modulus 2 and argument π/3, that is √ ω1 = 2cis(π/3) = 1 + 3i Furthermore, ω2 has modulus 2 and argument 5π/3, thus ω2 = 2cis(5π/3) = 1 −
√
3i
We know that, if −2 is a third root of z, we can obtain z by cubing −2. In other words, z = −8. We still avoid using any formulas, but we note that the modulus of this number is 8, therefore the fourth root will possess modulus 23/4 . Furthermore, the argument is θ/n where θ is the argument of z and n is the index of the root, thus for n = 4 we have argument π/4. So, we have the following roots, all with the same argument but separated by a factor of π/2: 23/4 cis(π/4) , 23/4 cis(−π/4) , 23/4 cis(3π/4) , 23/4 cis(−3π/4) This would suffice for a drawing, but we will also convert to a + bi form, yielding √ √ 2 2 ± ± 2 2 Similarly, for n = 5 we have modulus 23/5 and argument π/5. So, our roots are, all separated by an angle of 2π/5, as follows 23/5 cis(π/5) , 23/5 cis(3π/5) , 23/5 cis(π ) , 23/5 cis(−π/5) , 23/5 cis(−3π/5) In a + bi form that is approximately 1.2262 ± 0.8909i , −0.4684 ± 1.4415i , −1.5157 Note: This may be flawed; I will have to look at the work and interpretation again later.
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28.) Suppose n denotes a nonnegative integer. Determine the values of n such that zn = 1 possesses only real solutions. Defend your answer with sound mathematics. This question regarding roots of unity is deceivingly important and not entirely straightforward. Solving our equation we get that z = 11/n , so we use our nth root formula for 1, which has modulus 1 and argument 0, thus we have that the nth roots are √ 2πk 2πk n 1 = cos + i sin n n So, for answers in the reals we look at the image of the trigonometric functions sine and cosine, which, for both, is [−1, 1]. Thus, for a purely real answer we needn’t worry about the cosine (this will always be real) and instead turn our attention to sine, which is multiplied by i. Now, we must find which values of p satisfy the requirement (i · p) ∈ R. Now, we know that p = ni , n ∈ R satisfy this by definition of i and it also follows that p = 0 works because 0i = 0 ∈ R, but we must show that there exist no other p that meet the requirement. To do this is not immediately obvious but can be done quite nicely by cases. Suppose we have p where p ∈ R , p 6= 0 then we have some multiple of i, which is clearly not in the reals. If we have a complex, non-real p = a + bi such that a 6= 0 then we have a binomial to be distributed across, and the first value of that binomial must be complex because a ∈ R, thus ai ∈ C. So, all cases have been exhausted for p ∈ C, meaning that our only candidates are ni and 0. However, we must go back to the range of sine, which leaves us with but one possibility p = 0. So, we now go back to the main part of our problem and make the connection that p = sin
2πk n
= 0. Now, from the unit circle, we know that sin(φ) = 0 when φ = mπ , m ∈ Z. Because k and n are positive in our original, we only consider φ = mπ , m ∈ N. Now, upon investigation, we solve our equation in full and see that the nonnegative integer solutions are, all assuming n 6= 0: k = 0 and k = 2c1 c2 + c2 where n = 2c2 and c1 , c2 ∈ Z+ (this is a bit of a confusing argument which could perhaps be stated more simply).This is all very general and dense, but upon requiring that we must have only real solutions for a given n, our solution nicely simplifies. And so we can see that the only real roots of unity are ±1, which are achieved often but only exclusively for n = 1, 2 A somewhat clearer argument... Consider the geometry of roots. We know that all nth roots of unity lie on a circle of unit radius (i.e. r = 1). Therefore, for there to be a real root it must be that z = ±1, because any other real value would be interior or exterior to the unit disk. Now, consider the vector representation of a complex number z. Given the first root we have some number ω0 with modulus r and argument θ. Now, all other roots can be represented by the same vector, but with the argument adjusted by some factor of θ/n, where n is the same as in the nth root. Now, given that, because we can only achieve real numbers when the argument is kπ , k ∈ Z, we have that, because all roots with n ≥ 3 rotate by a factor smaller than 180◦ , they must attain at least some non-real values.
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31.) Discuss: A real number can have a complex nth root. Can a nonreal complex number have a real nth root? The answer to this is a bit of yes and no. As a pseudo-example consider the famous Euler’s identity, which, I would imagine, will be discussed extensively in this course. The formula’s derivation and proof is fairly simple with methods such as Taylor series, use of Euler’s formula, which relates the exponential and polar forms of a complex number, considerations regarding the complex exponential function ez = exp(z), where z ∈ C, et cetera. Here is the identity’s statement eiπ = −1 This simple but profound insight into mathematics also shows us an example of a number z ∈ C that has a real third root −1. However, despite this number having a nice expression in terms of i, it is actually a real number. An actual example of this phenomenon would be to consider the number z = i, then take its ith root. √i Interestingly, i = i1/i = i−i = eπ/2 ≈ 4.81, which, clearly, is a real number. Note here that this is actually a multi-valued function, so we have only provided one valid answer. That, however, is not quite what the question seems to be asking. Instead, it seems to be concerning only whole number roots given by √ √ θ + 2kπ θ + 2kπ n n z = r cos + i sin n n If this be the case, then for √ any complex number z = a + bi , b 6= 0 we have that the magnitude is some positive real number |z| = a2 + b2 and the argument is some real number on the interval [−π/2 , π/2] given by arctan(b/a). But, the argument cannot be zero, else the number would be real, i.e. b = 0. In order to have a real nth root we know that the i on the sine function must be eliminated. Because the sine function takes on only real values, the only way to eliminate the imaginary unit is to have sin( x ) = 0. But, for sine to be zero we must have that θ = πk , k ∈ Z, which would make the complex number real. Thus, the only way for the root of a number to be real is if the number itself is real. There is one exception, of course, because for any z ∈ C, z0 = 1. This proof is a bit informal, but it demonstrates the point. In conclusion, the nth root of a nonreal complex number z where n ∈ Z+ cannot have a real root. A better argument would be to argue that the reals are closed under multiplication, examining the equation z1/n = x , x ∈ R and translating it to x n = z, and finally noting that r n must be real via closure of the group (R, ·), so z is real.
21
Joseph Heavner Honors Complex Analysis Assignment 2 January 25, 2015
1.5 Sets of Points in the Complex Plane 1.) Sketch the graph of
|z − 4 + 3i | = 5
. Figure 1: Circle with radius 5 centered at 4 − 3i
3.) Sketch the graph of
|z + 3i | = 2
. Figure 2: Circle with radius 2 centered at −3i
5.) Sketch the graph of Re(z) = 5
. Figure 3: Vertical line at x = 5
1
7.) Sketch the graph of Im(z + 3i ) = 6 Unlike the previous problems, which were obvious enough as to call for no explanation, here we will write the equation in a more graph-able form. In particular, for z = a + bi I(z + 3i ) = 6 I( a − bi + 3i ) = 6
−b + 3 = 6 b = −3
. Figure 4: Horizontal line at y = −3
9.) Sketch the graph of
|Re(1 + iz)| = 3 |R(1 + i z¯)| = 3 |R(1 + i ( x − iy))| = 3 |1 + y | = 3 (1 + y )2 = 9 y = 2 , −4
. Figure 5: Horizontal lines at y = 2 and y = −4
2
11.) Sketch the graph of Re(z2 ) = 1 R(( a + bi )2 ) = 1 R( a2 + 2bi − b2 ) = 1 a2 − b2 = 1
. Figure 6: Hyperbola x2 − y2 = 1
13.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. Re(z) < −1 a.) yes
b.) no
c.) yes
d.) no
e.) yes
Some explanation is that any neighborhood in the region R is entirely contained (i.e. any e-ball is entirely contained), therefore it is open. Also, not all of the boundary points of R are contained in R, therefore it is closed. Because any two points z0 , z1 ∈ R can be connected by a collection of connected lines, the set is connected; by definition, an open, connected set is a domain. The set is not bounded, however, as the modulus of a point z0 ∈ R can become arbitrarily large (i.e. the set extends indefinitely towards x = −∞).
. Figure 7: Half-plane x < −1
3
15.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. Im(z) > 3 a.) yes
b.) no
c.) yes
d.) no
e.) yes
The justification here is identical to that in (13).
. Figure 8: Half-plane y > 3
17.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. 2 < Re(z − 1) < 4 To make this set easier to plot, we simplify the expression describing it a bit. 2 < R( z − 1) < 4 2 0 x 2 − y2 > 0 a.) yes
b.) no
c.) no
d.) no
e.) no
The justifications for (a), (b), and (d) are the same as in previous questions. This set is not connected because a line cannot possibly pass through the origin without leaving the set. Lack of connected-ness then implies that the set is not a domain.
. Figure 10: 3 < x < 5
21.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected.
|z − i | > 1 We know the general form of a circle from our text and the last homework assignment. a.) yes
b.) no
c.) yes
d.) no
e.) yes
The justification here is analogous to (13).
. Figure 11: The region outside the open disk x2 + (y − 1)2 < 1
5
23.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. 1 ≤ |z − 1 − i| < 2 This is an annulus like that described in our text. a.) no
b.) no
c.) no
d.) yes
e.) yes
For (b) and (e) the explanation is analogous to (13). This set is not open because, for the inner boundary, we could find a neighborhood that contains some point outside the set (in fact, every neighborhood on the boundary contains points not in the set). The set is not a domain, as it is not open. The set is bounded, because there exists a number r ∈ R : r > 0 such that |z| < r ∀z ∈ R. In particular, for this region, there exists no point in the region with modulus greater than 2 (with new coordinates such that the origin is the same as the center of the annulus).
. Figure 12: Circular annulus centered at (1,1) with inner boundary |z| = 1 and outer boundary |z| < 2
25.) Give the boundary points of the sets in Problems 13-24. I think this is straightforward enough that it requires no explanation. 13: x = −1 15: y = 3 17: x = 3, 5 23: |z − 1 − i | = 1 & |z − 1 − i | = 2
19: y = ± x
21: |z − i | = 1
27.) Sketch the set of points in the complex plane satisfying 0 ≤ arg(z) ≤ π/6 This question is most easily answered with a geometric explanation. Because the modulus of our complex number is not specified, we need only concern ourselves with the argument. We generally measure angles from the positive x-axis. Therefore, this region in polar coordinates would be (r, θ ≤ π/6). To me, this is already intuitive, the figure will be all z0 ∈ C in the area sweeped out by a ray of indefinite length with endpoint at the origin such that the angle between the ray and the positive x-axis is less than or equal to π/6 = 30◦ .
. Figure 13: A triangular region extending indefinitely 6
29.) Describe the shaded set in the given figure using arg(z) and an inequality. As a reference, below is the figure in question.
. Figure 14: Relevant image from the text With similar reasoning to (27) we know this is the concatenation of the regions 0 ≤ arg(z) ≤ 2π/3 and −2π/3 ≤ arg(z) ≤ 0. By combining the inequalities we get −2π/3 ≤ arg(z) ≤ 2π/3 or
|arg(z)| ≤
2π 3
31.) Solve the simultaneous equations
| z | = 2 , | z − 2| = 2 By the transitive property we have
q
| z | = | z − 2| q x 2 + y2 = ( x − 2)2 + y2 x 2 + y2 = ( x − 2)2 + y2 x2 + y2 = x2 − 4x + 4 + y2
−4x + 4 = 0 x=1 Now, we have to consider the second equality for a bit more information (i.e. a restriction).
( x − 2)2 + y2 = 4 x2 − 4x + 4 + y2 = 4 x2 − 4x + y2 = 0
(1)2 − 4(1) + y2 = 0 √ y=± 3 Now, we know that z = x + iy, therefore 1±
√
7
3i
33.) On page 28 we stated that if ρ1 > 0, then the set of points satisfying ρ1 < |z − z0 | is the exterior to the circle of radius ρ1 centered at z0 . In general, describe the set if ρ1 = 0. In particular, describe the set defined by |z + 2 − 5i | > 0 q ( x + 2)2 + ( y − 5)2 > 0
( x + 2)2 + ( y − 5)2 > 0 Now, let us consider some cases to figure out what this set is. Consider the case where x = −2, then we have that y 6= 5; similarly, if we let x 6= 2 then all y are valid; finally, y 6= 5 =⇒ ∀ x (note the loose use of notation), which is the same as the first case. Therefore, we have that our set describes the complex plane with the point (−2, 5) deleted, i.e. C − {−2 − 5i }. This can be thought of as the exterior of a circle of radius 0. To sum that up we have
{z : z 6= −2 − 5i } 37.) Suppose z0 and z1 are distinct points. Using only the concept of distance, describe in words the set of points z in the complex plane that satisfies
| z − z0 | = | z − z1 | The word ”concept” to me implies that one should avoid algebra in solving this problem. With that being said, this describes the set of points in the complex plane such that their distance to an arbitrary point (i.e. a variable) is equivalent. For instance, for the origin the points z0 = 1 and z1 = −i would satisfy this, as their moduli are equivalent, i.e. they are the same distance from the origin. In other words, this describes the set of all equivalent circles (for either a value of z or the values of z0 and z1 ); in particular, the intersecting points on the circles are described, so the result is a line passing through the midpoint of z0 and z1 and orthogonal to the segment joining z0 and z1 . 39.) Describe the shaded set in the given figure by the filling in the two blanks in set notation, using complex notation for equations and inequalities and one of the words and or or.
. Figure 15: Relevant image from the text This is the combination of the exterior of a circle centered at 4i with radius 4 and the interior of another circle with center (1, 3i ) and radius 3.
{z : |z − 4i | ≥ 4 ∪ |z − 1 − 3i | ≤ 3}
8
48.) Find a point ( x0 , y0 , z0 ) on the unit sphere that corresponds to the complex number 2 + 5i. Using parametric equations of lines as vectors, we can solve this quite easily. The line is as followed: x = 2t , y = 5t , z = 1 − t or (2t, 5t, 1 − t) Now, we know that ( x, y, z) must be one unit from the origin (by virtue of it lying on the unit sphere), so we use the distance formula as follows
(2t)2 + (5t)2 + (1 − t)2 = 1 30t2 − 2t = 0 1 t= 15 Therefore, our point is
(2/15 , 1/3 , 14/15)
50.) Express the coordinates of the point ( x0 , y0 , z0 ) on the unit sphere in Figure 1.5.10(b) in terms of the coordinates of the point ( a, b, 0) in the complex plane. Use these formulas to verify your answer to Problem 48. [ Hint: First show that all points on the line containing (0, 0, 1) and ( a, b, 0) are of the form (ta, tb, 1 − t).] I have changed my slightly more abstract geometric proof to the suggested one using parametrizations. We have that d = e = c = 0 and f = 1, so our parametrization is ( at, bt, 1 − t). now, similar to before, we will find a point, given our parametrization (ta, tb, 1 − t), that lies on the sphere and line. We do this as follows
( at)2 + (bt)2 + (1 − t)2 = 1 a2 t2 + b2 t2 + t2 − 2t = 0 t ( a2 t + b2 t + t − 2) = 0 2 t= 2 a + b2 + 1 By that fact we can now explicitly write our general mapping. 2Re(z) a = | z |2 +1
b=
c
=
2Im(z) | z |2 +1 | z |2 −1 | z |2 +1
or
2a 2b a2 + b2 − 1 , , a2 + b2 + 1 a2 + b2 + 1 a2 + b2 + 1
For z = 2 + 5i we have
2·2 2·5 22 + 52 − 1 , , 22 + 52 + 1 22 + 52 + 1 22 + 52 + 1
9
⇐⇒ (2/15 , 1/3 , 14/15)
2.1 Complex Functions 3.) Evaluate f (z) = loge |z| + iArg(z) at (a) 1
(b) 4i
(c) 1+i
a.) f (1) = ln(|1|) + iArg(1) = ln(1) + i (0) = 0 + 0 = 0 b.) f (4i ) = ln(|4i |) + iArg(4i ) = ln(4) +
π i 2
c.)
√ π 1 π f (1 + i ) = ln(|1 + i |) + iArg(i + 1) = ln( 2) + i = ln(2) + i 4 2 4 6.) Evaluate at (a) 2 − πi
f (z) = ez (b)
π 3i
(c) loge 2 −
5π 6 i
a.) f (2 − πi ) = e2−πi = b.)
e2 = − e2 eiπ
√ πi 1 2 π + i f ( i ) = e 3 = cos(π/3) + i sin(π/3) = 3 2 2
c.) f (loge 2 −
√ 5π 5πi 2 5π i ) = eloge 2− 6 i = 5πi = 2e− 6 = 2cis(−5π/6) = − 3 − i 6 e 6
9.) Find the real and imaginary parts u and v of f (z) = 6z − 5 + 9i Let z = x + iy, then f (z) = 6( x + iy) − 5 + 9i = 6x + 6iy − 5 + 9i = 6x − 5 + i (6y + 9) Thus, u = 6x − 5 v = 6y + 9
15.) Find the real and imaginary parts u and v of f (z) = e2z+1 Similar to in (9), we let z = x + iy, then e2x+2iy+1 = e2x · (cos(2iy + 1) + i sin(2iy + 1)) Now, we already have real and imaginary parts split, thus u = e2x cos(2y + 1) v = e2x sin(2y + 1) 10
17.) Find the real and imaginary parts u and v as a function of r and θ of f (z) = z Let z = x + iy, then z = x − iy = r (cos(θ ) + i sin(θ )) This then implies that u = r cos θ v = r sin θ 19.) Find the real and imaginary parts u and v as a function of r and θ of f ( z ) = z4 Let z = x + iy, then z4 = r4 (cis(4θ )) Where the result is by by de Moivre’s theorem. It then follows that u = r4 cos 4θ r4 sin 4θ
21.) Find the real and imaginary parts u and v as a function of r and θ of f (z) = ez Let z = x + iy, then ez = e x+iy = e x · eiy = er cos θ · eri sin θ = er cos θ [cos(r sin θ ) + i sin(r sin θ )] Therefore, u = er cos θ cos(r sin θ ) v = er cos θ sin(r sin θ )
24.) Find the natural domain of f (z) =
3z + 2i z3 + 4z2 + z
We factor the denominator as follows 3z + 2i 3z + 2i = z ( z ( z + 4) + 1) z(z2 + 4z + 1) √ √ It is now clear that the domain is all z such that z 6= 0, 3 − 2, − 3 − 2, where the last two factors are obtained by solving z2 + 4z = −1.
25.) Find the natural domain of f (z) =
iz | z − 1|
In order for this to be undefined we must have that |z − 1| = 0, therefore, for z = x + iy, we solve q | x + iy − 1| = 0 =⇒ ( x − 1)2 + y2 = 0 =⇒ x2 − 2x + 1 + y2 = 0 =⇒ x = 1 , y = 0 Therefore, the natural domain is all z such that z 6= 1.
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2.2 Complex Functions As Mappings 3.) Find the image S0 of the set S under f (z) = 3z S : Im(z) > 2 If the points in the domain must have imaginary part greater than two, then the image under multiplication by 3 means that the image’s imaginary part must be greater than 6. Thus, the image is the half plane with imaginary part greater than 6. In other words, S0 : Im(w) > 6
5.) Find the image S0 of the set S under f ( z ) = (1 + i ) z S : x = 2 We know that z must be of the form 2 + iy, so we find that f (2 + iy) = (1 + i )(2 + iy) = 2 + iy + 2i − y = (2 − y) + i (y + 2) Now, we have a system of equations to solve. Solving for y in terms of u we find that y = 2 − u, therefore the image is the line v = 4 − u 7.) Find the image S0 of the set S under f (z) = iz + 4 S : Im(z) ≤ 1 Break the domain into −∞ < x < ∞ , y ≤ 1. Our mapping is w = i ( x + iy) + 4 = (4 − y) + ix, therefore its parts are u = 4 − y , v = x. Now, we combine this to 4 − u ≤ 1 =⇒ u ≥ 3 and, of course, v is still a variable. Therefore, the image is the half plane S0 : Re(w) ≥ 3
9.) Find the image of the line y = 1 under w = z2 . For z = x + iy we have w = ( x + iy)2 = x2 − y2 + 2ixy, therefore u = x2 − y2 and v = 2xy, but for y = 1 2 we have u = x2 − 1 and v = 2x. So, we have that x = v/2, implying that u = v4 − 1. In other words, the image is the parabola 1 S 0 : u = v2 − 1 4 13.) Find the image of the line y = x under w = z2 . z = x + iy we have w = x2 − y2 + 2ixy, but x = y, therefore we have w = 2ix2 . Now, we know that x2 has domain of all real numbers, thus 2ix2 has domain C+ − R, in other words we have that the image is set of all z such that the y component varies indefinitely but the real part is zero, in other words this is the ray u = 0, with the restriction, due to the square, that 0 ≤ v < ∞. In short, the ray u = 0 , 0 ≤ v < ∞
12
15.) (a) Plot the parametric curve C given by z(t) and describe the curve in words, (b) find a parametrization of the image, C 0 , of C under the given complex mapping w = f (z), and (c) plot C 0 and describe this curve in words. z(t) = 2(1 − t) + it , 0 ≤ t ≤ 1; f (z) = 3z a.) Because z(0) = 2 and z(1) = i, we have that this curve C is the line segment from 2 to i. b.) w(t) = f (z(t)) = 3(2(1 − t) + it) = 6(1 − t) + 3it Thus, the parametrization is w(t) = 6(1 − t) + 3it , 0 ≤ t ≤ 1 c.) Similar to (a) but under the mapping in (b) we have that z0 = 6 and z1 = 3i, thus the image is the line segment from 6 to 3i.
. Figure 16: The mapping from the line segment from 2 to i to the line segment from 6 to 3i
17.) (a) Plot the parametric curve C given by z(t) and describe the curve in words, (b) find a parametrization of the image, C 0 , of C under the given complex mapping w = f (z), and (c) plot C 0 and describe this curve in words. z(t) = 1 + 2eit , 0 ≤ t ≤ 2π; f (z) = z + 1 − i a.) From our list of common parametrizations on page 57, we see this to be a circle, for it is of the form z(t) = z0 + reit , 0 ≤ t ≤ 2π. In particular, this is a circle of radius 2 centered at (1, 0).
. Figure 17: C, i.e. a circle of radius 2 centered at (1, 0) b.) This mapping is fairly straightforward; upon substitution of z into f (z) we have that w(t) = 2 − i + 2eit c.) This is then just another circle, but this time the set is a circle with center 2 − i and radius 2.
. Figure 18: C 0 , i.e. a circle of radius 2 centered at (2, −i )
13
21.) Use parametrizations to find the image, C 0 , of the curve C under f (z) = z3 ; C is the positive imaginary axis We have the parametrization z = it , 0 ≤ t ≤ ∞ for the curve, thus z = (it)3 = −it3 . From this, we have that, because t is a positive real number and only scales the imaginary unit, this forms a line on the negative imaginary axis. 23.) Use parametrizations to find the image, C 0 , of the curve C under f (z) = 1/z ; C : |z| = 2 If the modulus of an element of the domain of the mapping must be equal to 2, then the multiplicative inverse of that element must have modulus equal to the multiplicative inverse of the element’s modulus (i.e. 1/2). In other words, because the multiplicative identity z = 1 has modulus 1, for some z0 : |z0 | = 2, we must have that |z0−1 | = 1/2, as to ensure we reach the identity, which has modulus 1, under multiplication. In terms of parametrizations we have the parametrization z(t) = 2eit , 0 ≤ t < 2π. Furthermore, 1 1 f (z) = it = e−it , 0 ≤ t < 2π 2 2e This is the circle |w| = 1/2. 28.) Consider the parametrization z(t) = i (1 − t) + 3t, 0 ≤ t ≤ 1. (a) Describe in words this parametric curve. (b) What is the difference between the curve in part (a) and the curve defined by z(t) = 3(1 − t) + it , 0 ≤ t ≤ 1 (c) What is the difference between the curve in (a) and the curve define by the parametrization z(t) = 3 1 2 t + i (1 − 2 t ) , 0 ≤ t ≤ 2 (d) Find the parametrization of the line segment from 1 + 2i to 2 + i where the parameter satisfies 0 ≤ t ≤ 3. a.) By page 57, this is the line segment from i to 3. b.) Both are line segments between i and 3, but this one is from 3 to i rather than the reverse. However, line segments typically have no sense of directionality between points, so the two are practically the same. c.) Again, this is practically the same, but here the parameter takes on different values to produce the same geometry. d.) 2+i 1 (1 + 2i ) 1 − t + t 3 3 29.) Use the parametrizations to find the image of the circle |z − z0 | = R under the mapping f (z) = iz − 2. This mapping rotates by arg(i ) = π/2 and translates by −2. For a parametrization we have z(t) = z0 + Reit , 0 ≤ t ≤ 2π This is then mapped by iz − 2, meaning that w(t) = f (z(t)) = i (z0 + Reit ) − 2 , 0 ≤ t ≤ 2π Again, this describes the circle C shifted 2 units to the left (note that circles are geometrically invariant under rotation). 14
2.3 Linear Mappings 1) (a) Find the closed disk |z| ≤ 1 under the linear mapping f (z) and (b) represent the linear mapping with a sequence of plots. f (z) = z + 3i This is equivalent to shifting a disk of radius 1 by 3 units vertically, so the image is
| w − 3| ≤ 1
3.) (a) Find the closed disk |z| ≤ 1 under the linear mapping f (z) and (b) represent the linear mapping with a sequence of plots. f (z) = 3iz By equation (6) we have that f (z) = |3i | |3i3i z = 3(iz) + 0, thus the mapping rotates by arg(i ) = π/2 and magnifies by 3. Disks are invariant under rotation, but the magnifying factor still affects it. Thus, the function maps to a similar closed disk and is completely described as f : |z| ≤ 1 7→ |w| ≤ 3
. Figure 19: f : |z| ≤ 1 7→ |w| ≤ 3
5.) (a) Find the closed disk |z| ≤ 1 under the linear mapping f (z) and (b) represent the linear mapping with a sequence of plots. f (z) = 2z − i This is the mapping which magnifies the disk by a factor of 2 and shifts it vertically down i unit. In other words, the radius of our disk doubles and it’s center shifts by negative i.
|w + i | ≤ 2
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7.) (a) Find the image of the triangle with vertices 0, 1, and i under the linear mapping f (z) and (b) represent the linear mapping with a sequence of plots. f (z) = z + 2i The image of the triangle is not magnified or rotated, simply translated by 2i. Thus, the image is the triangle with vertices 2i, 1 + 2i, and 3i
. Figure 20: Triangle with vertices 2i, 1 + 2i, 3i
9.) (a) Find the image of the triangle with vertices 0, 1, and i under the linear mapping f (z) and (b) represent the linear mapping with a sequence of plots. f (z) = eiπ/4 z This mapping quite clearly rotates the triangle by π/4, so we just see what happens when we rotate the vertices, written in polar form, by this. In√particular, 0 maps to itself, 1 = e0 maps to eiπ/4 = √ 2/2 (1 + i ), and similarly i = eiπ 7−→ e5π/4 = 22 (1 − i ). So, we have the following result
√
√ √ √ 2 2 2 2 triangle with vertices 0 , + i, − + i 2 2 2 2 13.) Express the given linear mapping f (z) as a composition, rotation, magnification, and translation as in (6). Then describe the action of the linear mapping in words. f (z) = 3iz + 4 3i f (z) = |3i | z + 4 = 3(iz) + 4 |3i | So, the linear mapping rotates by arg i = π/2, magnifies by 3, and translates by 4. In other words, R(z) = eiπ/2 z , T (z) = z + 4 , M(z) = 3z.
15.) Express the given linear mapping f (z) as a composition, rotation, magnification, and translation as in (6). Then describe the action of the linear mapping in words.
√ 1 f (z) = − z + 1 − 3i 2 √ √ 1 −1/2 1 z + 1 − 3i = (−1z) + 1 − 3i f (z) = | − | 2 | − 1/2| 2 So, this mapping rotates by arg(−1) = π, magnifies by 21 , and translates by 1 −
16
√
3i.
17.) Find a linear mapping that maps S onto S0 when S is the triangle with vertices 0, 1, and 1 + i. S0 is the triangle with vertices 2i, 3i, and −1 + 3i. We know that our mapping must include T (z) = z + 2i, because 0 is invariant under rotation and magnification. With that in mind, we need a mapping with R(z) = π/2z, as that is how we can send 1 + 2i (which is 1 under translation) to the final 3i. Finally, we need no magnification as our last vertex maps correctly as is. So, we have f (z) = iz + 2i 19.) Find a linear mapping that maps S onto S0 when S is the imaginary axis and S0 is the line through the points i and 1 + 2i. I think of this as two parts. One is that we need to rotate our vertical line by −π/4. Second, we need to shift our line upward to have it pass through 1 + 2i rather than 1 + i, so we do so by 1. Thus,
√ f (z) = e
−iπ/4
z+1 =
√ ! 2 2 − i z+i 2 2
21.) Find the two different linear mappings that map the square with vertices 0, 1, 1 + i, and i, onto the square with vertices -1, 0, i, -1+i. The obvious mapping is translation by one. The less obvious one is to simply rotate our points by π/4, which actually has the same effect in this scenerio. So, our mappings are f (z) = z − 1 and g(z) = iz
23.) Consider the line segment parameterized by z(t) = z0 (1 − t) + z1 t , 0 ≤ t ≤ 1 (a) Find the parametrization of the image of the line segment under the translation T (z) = z + b , b 6= 0. Describe the image in words. (b) Find the parametrization of the image of the line segment under the rotation R(z) = az , | a| = 1. Describe the image in words. (c) Find the parametrization of the line segment under the magnification M(z) = az , a > 0. Describe the image in words. a.) w(t) = T (z(t)) = z0 (1 − t) + z1 t + b = z0 (1 − t) + z1 t + b(1 − t) + bt = (z0 + b)(1 − t) + (z1 + b)t This is the line segment between z0 + b and z1 + b (an analogous problem appears in our text). b.) w = R(z(t)) = a(z0 (1 − t) + z1 t) = ( az0 )(1 − t) + ( az1 )t is the line segment between az0 and az1 (again by (7) of 2.2 in our text). c.) w = M (z(t)) = a(z0 (1 − t) + z1 t) = ( az0 )(1 − t) + ( az1 )t This, virtually identical to the rotation previously described, is the line segment between az0 and az1 .
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2.4 Special Power Functions 1.) Find the image of the ray arg(z) = π/3 under the mapping w = z2 . Represent the mapping by drawing the set and its image. Our set C can be rewritten as z = reπ/3i . From here, finding the image of the set under the provided mapping is simple. 2π w = (reπ/3i )2 = Re2π/3i =⇒ arg(w) = 3 where R = r2 .
3.) Find the image of the line x = 3 under the mapping w = z2 . Represent the mapping by drawing the set and its image. Using equation (3) from page 75 we let k = 3, thus our solution is the parabola u = 9−
v2 36
−∞ < v < ∞
5.) Find the image of the line y = −1/4 under the mapping w = z2 . Represent the mapping by drawing the set and its image. From the general formula found on page 76 of our text we know the mapping to be of the form u = v2 − k2 . Here we have k = −1/4, therefore our image is 4k2 u=
v2 1 − (−1/4)2 =⇒ u = 4v2 − , v∈R 16 4(−1/4)2
9.) Find the image of the circular arc |z| = 1/2 , 0 ≤ arg(z) ≤ π under the mapping w = z2 . Represent the mapping by drawing the set and its image. Under the squaring function an arc has its modulus squared and both limits of its argument doubled, so we have that 1 |w| = , 0 ≤ arg(w) ≤ 2π 4
18
11.) Find the image of the triangle with vertices 0, 1, 1+i under the mapping w = z2 . Represent the mapping by drawing the set and its image. If we consider the three line segments composing the triangle to be separately mapped by w, then we can use know methods to find the image. In particular, the segment between 0 and 1 is mapped to itself (squaring the modulus and doubling the argument). The segment between 0 and 1 + i has argument π/4, therefore its image has argument π/2; similarly, the non-zero endpoint is squared, meaning that the segment maps to the segment from 0 to 2i. Finally, the segment from 1 to 1 + i lies on a vertical line, so, with k = 1, we have the segment mapping to u = 1 − v2 /4; however, for the bounds we look at our two points, which shows us that 0 ≤ v ≤ 2. Therefore, the image is the combination of the line segment from 0 to 1, the line segment from 0 to 2i, and the 2 parabolic arc u = 1 − v4 , 0 ≤ v ≤ 2.
. Figure 21: The mapping f : C → C described in (11)
15.) Find the image of the ray arg(z) = π/3 under f (z) = 2z2 + 1 − i. Let us consider this as a composition. In particular, f (z) = (h ◦ g)(z) where h(z) = 2z + 1 − i and g(z) = z2 . Under squaring arg(z) = π/3 maps to arg(w) = 2π/3. The linear transformation then simply translates and magnifies. The ray is invariant under magnification, but translation by 1 − i makes the image a ray from 1 − i making angle 2π/3 with the line y = 1. Because zero cannot possibly have argument π/3 it is not in the original set, thus our image excludes 1 − i. Finally, if we consider √ √ the point 12 + 23 i in our domain, then we see that it maps to 3i − i. Therefore, our mapping is the ray √ emanating from 1i, containing ( 3 − 1)i, and excluding the point 1 − i. 17.) Find the image of the line x = 2 under f (z) = iz2 − 3. 2
v We use equation (3) from page 75, with k = 2. This shows that the squaring function maps to u = 4 − 16 , but this is then rotated by arg(i ) = π/2 and translated by 3. Translation affects our imaginary part v, but rotation by that angle switches real and imaginary parts, making our image the parabola
v = 4−
1 ( u + 3)2 16
19.) Find the image of the circular arc |z| = 2 , 0 ≤ arg(z) ≤ π/2 under f (z) = 1/4eiπ/4 z2 . As with (15) we have a composition f (z) = (h ◦ g)(z), this time with h(z) = 41 eiπ/4 z and g(z) = z2 . The squaring function maps our circular arc to another such arch with |w| = 4 , 0 ≤ arg(w) ≤ π. The linear function rotates by π/4 and translates by 0; therefore, the circular arc maps again onto the arc |w| = 4 , π/4 ≤ arg(w) ≤ 5π/4. Finally, magnification brings about the final result. The image is the circular arc |w| = 1 , π/4 ≤ arg(w) ≤ 5π/4
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25.) Use (14) to find the value of z1/2 at z = −i. By (14) we have
(−i )1/2 =
q
| − i |eiArg(−i)/2 = e(−π/4) i = cis(−π/4) =
√
2/2 −
√
2/2i
27.) Use (14) to find the value of z1/3 at z = −1. Similar to (25) we have
(−1)1/3 =
q
| − 1|eiArg(−1)/3 = eπi/3 =
29.) Use (14) to find the value of z1/4 at z = −1 +
√
√ 1 3 + i 2 2
3i.
Again,
(−1 +
√
3i )1/4 =
q 4
|−1+
√
3i | eiArg(−1+
√
3i )/4
=
√ 4
2eiπ/6 =
√ 4
√ 2
3 1 + i 2 2
!
=
1√ 1√ 4 4 18 + 2i 2 2
31.) Find the image of the ray arg(z) = π/4 under the mapping w = z1/2 . Represent the mapping by drawing the set and its image. If the argument of z is π/4, then z must be of the form reiπ/4 , thus we have the following from our formula z1/2 = reiπ/8 = r (.925 + .383i ) Now, we see that this is equivalent to a linear mapping in which the complex number .925 + .383i is magnified by r for varying r. The image is thus the ray r (.925 + .383i ), or, in more simple terms, the ray arg(w) = π/8.
33.) Find the image of the positive imaginary axis under the mapping w = z1/2 . Represent the mapping by drawing the set and its image. This set is equivalent to arg(z) = π/2. Therefore, the image is the ray arg(z) =
20
π 4
35.) Find the image of the arc |z| = 9 , −π/2 ≤ arg(z) ≤ π under the mapping w = z1/2 . Represent the mapping by drawing the set and its image. Because the principal square root function roots the modulus and halves the argument of any number, we have that the image is simply the circular arc |w| = 3 , −π/4 ≤ arg(w) ≤ π/2.
39.) Find the image of the region shown in Figure 2.4.22 under the function w = z1/2 . (Be careful near the negative real axis!) This figure is boundary formed by the union of the ray arg(z) = π/2 and parabola x = 4 − 1/16y2 . With that being said, the principal square root function halves all arguments, so the ray maps to arg(z) = π/4 (note that this is equivalent to u = v). By the inverse relationship of z2 and z1/2 , as well as the formula for the mapping of a line onto a parabola under the squaring function (in this case k = 2), we know the parabola to map to u = 2. Now, let us consider a point within our domain. For instance, 3 + 4i. This, upon squaring, yields −7 + 24i. Now, we may completely specify our region as the set bounded by the lines u = v and u = 2 containing the point 3 + 4i.
. Figure 22: The union of the ray arg(z) = π/2 and parabola x = 4 − 1/16y2 .
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Joseph Heavner Honors Complex Analysis (Continued) Chapter 4 July 11, 2015
4.1 Exponential and Logarithmic Functions 3.) Find the derivative of f (z) = eiz − e−iz . d dz
eiz − e−iz =
d d iz −iz ) dz ( e ) − dz ( e
d d = eiz dz (iz) − e−iz dz (−iz) = ieiz + ie−iz
7.) Write arg ei(z+z) in terms of x and y. By (5) we have the argument is simply I(i (z + z)), so we simply expand this to get the desired result. I(i (z + z)) = I(i (( x + iy) + ( x − iy))) = I(2ix ) = 2x But, we must recall periodicity, which provides the final solution 2x + 2nπ , n = 0, ±1, ±2, . . . 15.) Find the image of the line y = −2 under the exponential mapping. Horizontal lines map to rays, in particular the line y = c 7→ Arg(w) = c, thus the solution is simply Arg(w) = −2
23.) Find all complex values of ln(−2 + 2i ). The complex natural logarithm is simply ln(r ) + iθ, and we have that r = so the logarithm is ln(2
3/2
)+i
3π + 2nπ 4
=
√
8 = e3/2 and θ =
3π 4
+ 2nπ,
3 (8n + 3)π ln(2) + i , n = 0, ±1, ±2, . . . 2 4
27.) Find the principal value of Ln(6 − 6i ) in a + bi form. √ Here we restrict −π < θ ≤ π and have r = 72 and θ = −π/4, so
√ iπ 1 π Ln(6 − 6i ) = ln( 72) − = ln(72) − i 4 2 4 35.) Find all z ∈ C satisfying ez−1 = −ie3 . This implies that z − 1 = ln(−ie3 ), which means z = 1 + ln(−ie3 ). We simplify using Definition 4.1.2: π (4n − 1)π (4n − 1)π 1 + ln(e3 ) + i − + 2nπ = 1 + 3 ln(e) + i = 4+ i , n = 0, ±1, ±2, . . . 2 2 2
1
43.) Find the image of the circle |z| = 4 under the mapping w = Ln(z). Circles map to vertical lines under this mapping, in particular this maps to the vertical line segment u = ln(4) , −π < v ≤ π 47.) Use (1) to prove that ez1 /ez2 = ez1 −z2 . This and similar properties are simple enough to prove. ez1 /ez2 =
e x1 (cos(y1 ) + i sin(y1 )) e x2 (cos(y2 ) + i sin(y2 ))
But we know division of complex numbers amounts to dividing moduli and subtracting arguments, so we have ez1 /ez2 = e x1 − x2 [cos(y1 − y2 ) + i sin(y1 − y2 )] And this gives us the desired ez1 −z2 .
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4.2 Complex Powers 3.) Find all values of (1 + i )1−i . √ |z| = 2 and θ = π/4 + 2nπ, thus
√ 1 (8n + 1)π ln(1 + i ) = ln( 2) + i (π/4 + 2nπ ) = ln(2) + i , n = 0, ±1, ±2, . . . 2 4 So, we have 1
(1 + i )1−i = e(1−i) ln(1+i) = e(1−i)( 2 ln(2)+ Noting that eln(2)/2 =
(1 + i )
1− i
√
=
(8n+1)π i) 4
= e(ln(2))/2+(8n+1)π/4+i[(8n+1)π/4−(ln(2))/2]
2 yields the simplified (a misnomer here) solution
√
2 exp
(8n + 1)π +i 4
(8n + 1)π ln(2) − 4 2
, n = 0, ±1, ±2, . . .
7.) Find the principal value of (−1)3i . This is less messy than the last problem. |z| = 1 and Arg(z) = π, so Ln(−1) = ln(1) + iπ = iπ. Thus,
(−1)3i = e3iLn(−1) = e−3π 15.) Find the derivative of z3/2 on the domain |z| > 0, −π < arg(z) < π at z = 1 + i. Since 1 + i is in the function’s usual domain we have that f 0 ( z ) | z =1+ i =
3 1/2 3 z |z=1+i = (1 + i )1/2 2 2
If one were so inclined, one could also find that
(z
√
1+i =
√ 4
√ 3 4 2 πi/8 ) = e 2
3/2 0
3
πi
2e 8 , so
4.3 Trigonometric and Hyperbolic Functions π 2
7.) Express sec
sec
π 2
−i =
− i in terms of a + bi. 1
cos
π 2
−i
=
1 1 = = −icsch(1) cos(π/2) cosh(−1) − i sin(π/2) sinh(−1) i sinh(1)
11.) Find all z ∈ C such that sin z = cos z. Using the definition of the trigonometric functions in terms of the exponential (and skipping a few steps) eiz + e−iz eiz − e−iz = 2i 2 e2iz − 1 = ie2iz + i
(eiz )2 = i Taking the square root of i yields eiπ/4 and e5iπ/4 , so we solve two equations, setting both of those equal to eiz . To save some time, the work will only be show for the first case. eiz = eiπ/4 iz = ln(1) + i
π
+ 2πk
4 (1 + 8k)π z= , k = 0, ±1, ±2, . . . 4 7π π 9π z = ··· ,− , , ,··· 4 4 4
For the second case we have a similar solution: z = ··· ,−
3π 5π 13π , , ,··· 4 4 4
Thus, z=
(4n + 1)π , n = 0, ±1, ±2, . . . 4
15.) Verify that cos z = cos z ∀z ∈ C. cos z = cos( x ) cosh(−y) − i sin( x ) sinh(−y)
=
cos( x ) cosh(y) + i sin( x ) sinh(y) cos( x ) cosh(y) − i sin( x ) sinh(y)
= cos z
19.) Find
d dz
d dz
1 z tan( ) z
z tan( 1z ) .
1 = z · tan z
0
0 1 1 1 1 1 2 1 2 1 + tan = z sec + tan = − sec + tan z z z z z z z 4
23.) Express cosh(1 +
π 6 i)
in a + bi form.
cosh(1 +
π π i ) = cos i 1 + i 6 π 6 = cos − + i 6 π π = cos − cosh(1) − i sin − sinh(1) 6 6 √ 3 i = cosh(1) + sinh(1) 2 2
27.) Find all z ∈ C such that sinh z = cosh z. By definition
ez
ez − e−z ez + e−z = 2 2 z z z −z = e e + e−z e −e
−1 = 1 Therefore, no solutions exist. 31.) Verify that | sinh z|2 = sinh2 x + sin2 y.
| sinh z|2 = | − i sin(iz)|2 = | sin(−y + ix )|2 = sin2 (−y) + sinh2 ( x ) = (− sin y)2 + sinh2 x = sin2 (y) + sinh2 ( x )
35.) Find the derivative of tanh(iz − 2).
(tanh(iz − 2))0 = sech2 (iz − 2) · (iz − 2)0 = sech2 (iz − 2) · i = isech2 (iz − 2)
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4.4 Inverse Trigonometric and Hyperbolic Functions 7.) Find all values of sinh−1 i.
π (4n + 1)π sinh−1 (i ) = ln i + (−1 + 1)1/2 = ln(1) + i ( + 2nπ ) = i , n = 0, ±1, ±2, . . . 2 2
√ 15.) Use the branch reiθ/2 , −2π < θ < 0 and the principal branch of ln z to (a) find the value of cosh−1 z at z = −i, and (b) find the derivative of cosh−1 z at that point. (a) cosh−1 (−i ) = ln −i + (−1 − 1)1/2 = ln −i + (−2)1/2 √ = ln −i + 2e−iπ/2 √ = ln −( 2 + 1)i √ π = ln 2+1 − i 2 (b)
cosh−1 (z)
0
1 |z=−i − 1)1/2 1 = (−2)1/2 1 = √ iπ 2e− 2 √ 2 i = 2
|z=−i =
( z2
19.) Use implicit differentiation to derive formula (8) for the derivative of the of the inverse cosine. Note that w = cos−1 (z) ⇔ z = cos(w), so we implicitly differentiate to get 1 = − sin z · But, 1 = sin2 w + cos2 w = sin2 w + z2 , so
√
dw dw −1 ⇔ = dz dz sin w
1 − z2 = sin w. Thus,
dw −1 −1 = = √ dz sin w 1 − z2
6
Special Topic: The Lambert W Function While we are on the topic of the exponential function and its inverse, the logarithmic function, it is worth briefly discussing the famous Lambert W. This function can be defined by the following z = W ( z ) eW ( z ) , z ∈ C As with most complex functions, this is multi-valued. However, we can restrict the function to be single-valued. While this function generally comes in play in combinatorics or applied settings, we shall look at a single example from the simple setting of exponential equations. This is a particularly significant equation, one that is not only common but also instructive. Example Solve the equation x x = z. We begin with logarithms xx = z x ln x = ln z e
ln x
ln x = ln z
Now, we invoke the W function. eln x ln x = ln z ln x = W (ln z) x = eW (ln z)
It is worth looking into this further, if only just for entertainment value. http://goo.gl/HW5kMG provides a good image of the function’s Riemann surface as well as some examples, http://goo.gl/ cBEldp is a good technical exposition, and finally https://goo.gl/RO4qlp provides an extended, somewhat technical introduction as well as some history.
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