Complex number hsc nsw extension 2 maths
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Complex number hsc nsw extension 2 maths...
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NORMANHURST BOYS HIGH SCHOOL
MATHEMATICS (EXTENSION 2)
Notes and exercises: With references to
4
Complex numbers
Name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Initial version by H. Lam, August 2012. Updated April 4, 2016. Various corrections by students & members of the Department of Mathematics at Normanhurst Boys High School and Normanhurst Boys High School. Acknowledgements Pictograms in this document are a derivative of the work originally by Freepik at http://www.flaticon.com, used under CC BY 2.0.
Symbols used !
Beware! Heed warning.
2
Mathematics content.
3
Mathematics Extension 1 content.
L
Literacy: note new word/phrase.
M
Formulae/facts must be memorised.
R the set of real numbers C the set of complex numbers ∀ for all
Gentle reminder • For a thorough understanding of the topic, every blank space/example question in this handout is to be completed! • Additional questions from the selected texts will be completed at the discretion of your teacher. • Remember to copy the question into your exercise book!
Contents 1 A new number system 1.1 Review of number systems . . . . . . . 1.2 Rotation . . . . . . . . . . . . . . . . . 1.3 The “imaginary” numbers . . . . . . . 1.4 Basic operations with complex numbers 1.4.1 Addition . . . . . . . . . . . . . 1.4.2 Multiplication . . . . . . . . . . 1.4.3 Complex conjugate pairs . . . . 1.5 Properties of complex numbers . . . . 1.5.1 Equality . . . . . . . . . . . . . 1.5.2 Solutions to equations . . . . .
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5 5 6 7 9 9 9 10 13 13 13
numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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15 15 15 16 16 17 19 19 19 20 22 24 27
3 Locus problems 3.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 32 34
4 Past HSC questions 4.1 2001 Extension 2 HSC 4.2 2002 Extension 2 HSC 4.3 2003 Extension 2 HSC 4.4 2004 Extension 2 HSC 4.5 2005 Extension 2 HSC 4.6 2006 Extension 2 HSC 4.7 2007 Extension 2 HSC 4.8 2008 Extension 2 HSC
40 40 42 43 44 45 47 47 49
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2 Further arithmetic & algebra of complex 2.1 Vector representation . . . . . . . . . . . 2.1.1 Equivalence . . . . . . . . . . . . 2.1.2 Addition . . . . . . . . . . . . . . 2.1.3 Subtraction . . . . . . . . . . . . 2.1.4 Scalar multiplication . . . . . . . 2.2 Modulus/argument of a complex number 2.2.1 Natural ordering . . . . . . . . . 2.2.2 Modulus . . . . . . . . . . . . . . 2.2.3 (Principal) Argument . . . . . . . 2.2.4 Triangle inequality . . . . . . . . 2.3 Mod-arg & polar form . . . . . . . . . . 2.3.1 Properties . . . . . . . . . . . . .
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4
Contents –
4.9 4.10 4.11 4.12 4.13 4.14 4.15
2009 2010 2011 2012 2013 2014 2015
References
Extension Extension Extension Extension Extension Extension Extension
2 2 2 2 2 2 2
HSC HSC HSC HSC HSC HSC HSC
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50 51 51 52 54 56 57 58
NORMANHURST BOYS’ HIGH SCHOOL
Section 1 A new number system 1.1 Review of number systems • . . . . . . . . . . . . . . . . . . numbers. N = {1, 2, 3 · · · } Example 1
Solve x + 1 = 5 and x + 3 = 0 over N.
Answer: x = 4, no solution
• . . . . . . . . . . . . . . . . . . . Z = {· · · , −3, −2, −1, 0, 1, 2, 3, · · · } Example 2
Solve x + 3 = 0 and 2x + 4 = 7 over Z.
• . . . . . . . . . . . . . . . . . . . . numbers. Q = Example 3
p : p, q ∈ Z, q 6= 0 q
Solve 2x + 4 = 7 and x2 − 2 = 0 over Q.
Answer: x = −3, no solution
Answer: x =
3 , 2
no solution
• . . . . . . . . . . numbers. R Example 4
2
Solve x − 2 = 0 and x2 + 5 = 0 over R.
• N⊆Z⊆Q⊆R
5
√ Answer: x = ± 2, no solution
6
A new number system – Rotation
√ 5 4 e7 1.1 5π 2
log3 8 4 7
11 23
5.45
−8 −7
√ −1+ 50.4453 2
−9 2
1
Q Z
−10
sin 1.25
√
3 5
3 N 6 −4 4 5 −6 −5 1 2
R 2
75 9 2
73
6 16
cosh 7
1.2 Rotation • From x = 1, go to x = −1 by rotating π radians in the usual direction. – Multiply 1 by −1 to obtain −1 corresponds to rotating by . . radians. • Stop halfway whilst rotating? Quarter of way whilst rotating?
−2
−1
0
1
2
NORMANHURST BOYS’ HIGH SCHOOL
A new number system – The “imaginary” numbers
7
1.3 The “imaginary” numbers Definition 1 The imaginary number i to be the “quantity” to multiply with a real number when rotating anti-clockwise by π2 about x = 0. Imaginary number
• “Jump off” the real number line.
Definition 2 The imaginary number i has property such that i × i = i2 = −1 • Why?
Definition 3 The set of all imaginary numbers, called the complex numbers, is defined to be C = {z : z = x + iy; x, y ∈ R}
Example 5 Find the values of i2 , i3 , i4 and i5 . • i2 = . . . . . . . . . .
NORMANHURST BOYS’ HIGH SCHOOL
• i3 = . . . . . . . . . .
• i4 = . . . . . . . . . .
• i5 = . . . . . . . . . .
8
A new number system – The “imaginary” numbers
Definition 4 Complex number A complex number z has real and imaginary parts and is defined by z = x + iy. • The real part of z: Re(z) = x. • The imaginary part of z: Im(z) = y.
• Treat real and imaginary parts as . . . . . . . . . . . . . . . . . . . . . . . . . . . . of a complex number. • z = x + iy is known as . . . . . . . . . . . . . . . . . . . . . . form. • Plot on A . . . . . . . . . . . . . d . . . . . . . . . . . . . . . . , similar to plotting points coordinate geometry. Example 6 On the following diagram, plot the location of: • z1 = 3 + 4i.
• z2 = 2 − i.
• z4 = − 21 + 32 i.
• z3 = −1 − 3i. Im 4 3 2 1
Re −4
−3
−2
−1 −1
1
2
3
4
−2 −3 −4
NORMANHURST BOYS’ HIGH SCHOOL
A new number system – Basic operations with complex numbers
9
1.4 Basic operations with complex numbers Example 7 Find the value of √ √ 1. 2+ 3 − 5−4 3
1.4.1
2.
2+
√ √ 3 5−4 3
Addition
• Operations similar to surds (group rational parts with rational parts, irrational parts with irrational parts). • Group . . . . . . . . parts with . . . . . . . . parts • Group . . . . . . . . . . . . . . . . . . . . . . . parts with . . . . . . . . . . . . . . . . . . . . . . . parts. 1.4.2
Multiplication
• Use distributive law. • Beware that i2 = . . . . , which becomes . . . . . . . . Example 8 If z1 = 2 + 3i and z2 = −1 + 5i, find the value of (a)
z1 + z2
(b)
z1 − z2
(c)
3z1
(d)
Example 9 Find z ∈ C such that Re(z) = 2 and z 2 is imaginary.
NORMANHURST BOYS’ HIGH SCHOOL
3iz1
(e)
z1 z2
10
A new number system – Basic operations with complex numbers
1.4.3
Complex conjugate pairs Definition 5
If z = x + iy, then its complex conjugate is denoted z such that z = x − iy √ • Analogous to conjugate surds, where the conjugate of a + b c is . . . . . . . . . . . . . . . . • Geometrically, Im b
z = x + iy
Re
Example 10 If z1 = 2 + i and z2 = 1 − 3i, evaluate in Cartesian form: (a)
z1 + z2
(c)
−z2
(e)
(b)
z2 − z1
(d)
z1
(f)
z1 − z2 1 z2
NORMANHURST BOYS’ HIGH SCHOOL
A new number system – Basic operations with complex numbers
Example 11 Find the square roots of −3 + 4i in Cartesian form.
NORMANHURST BOYS’ HIGH SCHOOL
11
12
A new number system – Basic operations with complex numbers
History Gerolamo Cardano (1501-1576), Mathematician (gambler and chess player!), published solutions to the cubic ax3 + bx + c = 0 in Ars Magna. Cardano was one of the first to acknowledge the existence of imaginary numbers. Given during the Renaissance, negative numbers were treated suspiciously, imaginary numbers would have been almost heretical.
Cardano did not avoid (as most contemporaries did) nor did he immediately provide solutions to these imaginary numbers (possibly 200 years away). With the equations containing complex conjugate pairs, Cardano multiplied them together and obtained real numbers: √ Putting aside the mental tortures involved, multiply 5 + −15 with √ 5 − −15, making 25 − (−15), which is −15. Hence the product is 40. √ Cardano, remarked in another work, that −9 is neither +3 or −3, but some “obscure sort of thing”. Source: • Wikipedia (http://en.wikipedia.org/wiki/Gerolamo Cardano) • Complex and unpredictable Cardano, Artur Ekert, Mathematical Institute, University of Oxford, United Kingdom (http://www.arturekert.org/Site/Varia files/NewCardano.pdf)
Further exercises • Lee (2006, Ex 2.3)
• Sadler and Ward (2014, Ex 1A) NORMANHURST BOYS’ HIGH SCHOOL
A new number system – Properties of complex numbers
13
1.5 Properties of complex numbers 1.5.1
Equality Definition 6
Two complex numbers z1 and z2 are equal iff the real and imaginary parts are equal.
Proof
• Let z1 = a + ib, z2 = c + id • If z1 = z2 , then . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , or • ....................................................
1.5.2
Solutions to equations Example 12
Solve z 2 + 1 = 0 for z ∈ C.
Answer: z = ±i
Example 13 Solve z 2 + 2z + 10 = 0 for z ∈ C.
NORMANHURST BOYS’ HIGH SCHOOL
Answer: z = −1 ± 3i
14
A new number system – Properties of complex numbers
Example 14 Solve 2z 2 + (1 − i)z + (1 − i) = 0 for z ∈ C.
Answer: z = i, z = − 21 − 12 i
Observations
• Equations with . . . . . . . . coefficients will have . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . roots. • Equations with . . . . . . . . . . . . . . . . . . . coefficients do not necessarily have . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . roots.
Further exercises • Sadler and Ward (2014, Ex 1B) • Fitzpatrick (1991, Ex 31(a), (b), (c)) • Lee (2006, Ex 2.1, 2.2) • Arnold and Arnold (2000, Ex 2.1) NORMANHURST BOYS’ HIGH SCHOOL
Section 2 Further arithmetic & algebra of complex numbers 2.1 Vector representation 2.1.1
Equivalence Definition 7
Two vectors p and q on the Argand diagram are equal iff both • Modulus ( . . . . . . . . . . . . . . . . . . . . . . . . ), and • Argument ( . . . . . . . . . . . . . . . . . . . . ) are equal.
• The starting point ( . . . . . . . ) is irrelevant for a vector.
15
16
Further arithmetic & algebra of complex numbers – Vector representation
2.1.2
Addition
• Place vectors, head-to-tail. Im p
q p+q
Re
O q
• Parallelogram of vectors when adding two vectors. Example 15 If z = −3 + 2i and w = 2 + 4i, draw z + w on the Argand diagram. Im Im 6
6
4
4
2
2 Re
−3 −2 −1 −2 2.1.3
1
2
3
Re −3 −2 −1 −2
1
2
3
Subtraction
• For z1 − z2 , add −z2 to z1 . Im −q p + (−q)
p−q
p
Re p+q
O q
NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Vector representation
17
• Alternatively, “what vector get you to z1 from z2 ?” • Quick & easy parallelogram of vectors: 1.
z1 + z2 starts from tails of z1 & z2 ,
2.
z2 − z1 starts from head of z1 , goes to head of z2 .
Example 16 If z = 3 − 2i and w = 2 − 5i, draw z − w on the Argand diagram Im Im 3 2 1
3 2 1 Re
−1−1 −2 −3 −4 −5
2.1.4
1
2
3
Re −1−1 −2 −3 −4 −5
1
2
3
Scalar multiplication
• For kz1 where k ∈ R, stretch z1 by factor of k. • If k < 0, direction of new vector is opposite to original vector. Example 17 If z = 1 + 2i, draw 3z and −2z on separate Argand diagrams. Im Im 6 5 4 3 2 1 −3 −2 −1 −2 −3 −4
NORMANHURST BOYS’ HIGH SCHOOL
b
6 5 4 3 2 1
z Re
1 2 3 4
−3 −2 −1 −2 −3 −4
b
z Re
1 2 3 4
18
Further arithmetic & algebra of complex numbers – Vector representation
Example 18
√ √ On the Argand diagram, the complex numbers 0, 1 + i 3, 3 + i, and z form a rhombus. Im z [2011 HSC Q2]
b
√ 1+i 3 b
θ √ b
b
O
3+i Re
(i)
Find z in the form a + ib, where a and b are real numbers.
(ii)
An interior angle, θ, of the rhombus is marked on the diagram. Find the value of θ. √ √ Answer: z = ( 3 + 1) + i( 3 + 1), θ =
5π 6
Further exercises • Sadler and Ward (2014, Ex 1C, 1E) NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number
2.2 Modulus/argument of a complex number 2.2.1
Natural ordering
• N, Z, Q, R are well ordered : – 6 ... 1
–
22 7
... π
–
5 4
...
5 7
• However: – 6 + 4i . . . . . . 3 + 2i
– 3 − 3i . . . . . . 2 + i
• Natural ordering does not exist with complex numbers. 2.2.2
Modulus Definition 8
The modulus of a complex number, denoted |z| (where z = x + iy) is the magnitude ( . . . . . . . . . . . . . . ) of the vector from O to z on the Argand diagram. Im z = x + iy
|z
|
b
y
Re x i.e. |z| =
NORMANHURST BOYS’ HIGH SCHOOL
p
x2 + y 2
19
20
Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number
2.2.3
(Principal) Argument Definition 9
The argument of a complex number, measured in . . . . . . . . . . . . . . . . . , is denoted arg(z) (where z = x + iy) is the . . . . . . . . . . . . that the vector from O to z makes with the positive real axis on the Argand diagram, with angles increasing in the anticlockwise direction. Im z = x + iy b
arg(z)
i.e. arg(z) = tan−1
Re
y x
• Duplicate argument(s)? Example 19 Evaluate arg(z), where z = 1 + i.
Definition 10 The principal argument of a complex number, denoted Arg(z) lies within the domain −π < Arg(z) ≤ π. Important note • The principal argument is generally quoted henceforth. • Be aware of the quadrant which z lies. Inputting tan−1 give an erroneous result.
y x
on the calculator may
• The complex number z = 0 + 0i has no argument defined. NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number
21
Example 20 Find the modulus and principal argument of the following: (a)
2 + 2i
√ Answer: modulus: 2 2, argument
(b)
√ −1 − i 3
Answer: modulus: 2, argument − 2π 3
π 4
Example 21 [2011 HSC Q2]
Let w = 2 − 3i and z = 3 + 4i.
(a)
Find w + z.
(b)
Find |w|. w in the form a + ib, where a, b ∈ R. Express z
(c)
NORMANHURST BOYS’ HIGH SCHOOL
Answer: 5 + i Answer:
√
6 Answer: − 25 −
13
17 i 25
22
Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number
2.2.4
Triangle inequality Theorem 1
For every complex number z1 and z2 , |z1 + z2 | ≤ |z1 | + |z2 | Proof
1
Steps 1.
Let p and q (with P and Q being the head of the arrow) represent the complex numbers z1 and z2 respectively, p + q with R being the head of the arrow.
2.
On the Argand diagram: Im
Re
3.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iff O, P and Q are collinear (which implies OP k OQ k OR) • Conclusion: z1 = kz2 , where k ∈ R as vectors are parallel.
4.
Otherwise, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.
Hence, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Never attempt to prove this algebraically!
NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Modulus/argument of a complex number
Example 22 If z1 = 3 + 4i and |z2 | = 13, find the greatest value of |z1 + z2 |. If |z1 + z2 | is at its greatest value, find the value of z2 in Cartesian form. Answer: |z1 + z2 | = 18 at its greatest; z2 =
39 5
+
52 i 5
Further exercises • Lee (2006, Ex 2.6 Q1-7) • Patel (2004, Ex 4K) NORMANHURST BOYS’ HIGH SCHOOL
• Arnold and Arnold (2000, Ex 2.3)
23
24
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
2.3 Modulus-argument & polar form Definition 11 The modulus-argument form of a complex number z is Im
z = x + iy
|z
|
z = x + iy (Cartesian form) = |z| cos θ + i |z| sin θ (Mod-arg form) = |z| (cos θ + i sin θ) = r (cos θ + i sin θ)
θ
where Arg(z) = θ.
y = |z| sin θ
b
Re
x = |z| cos θ
• . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . often abbreviated2 to z = r cis θ. • Better to abbreviate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . to z = reiθ (for reasons that will be made obvious later)
Definition 12 Polar form: Euler’s formula eiθ = cos θ + i sin θ where e ≈ 2.71828 · · · (Leonhard Euler, 1707-1783. http://en.wikipedia.org/wiki/Leonhard Euler) (All index laws in R also apply to the complex exponential).
2
cis θ does very little to assist your understanding of the rules for multiplying complex numbers!
NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
Example 23 √ Write z = 2 cos 3π + i sin 3π in Cartesian form. 4 4
25
Answer: z = −1 + i
Example 24 Write z = −2 − 2i in polar form.
√ 3π Answer: z = 2 2e−i 4
Example 25 10 in polar form, and hence write in simplest Cartesian form. Write z = √ 3+i Answer: z = 5e−
iπ 6
=
√ 5 3 2
− 52 i
Example 26 Write z = 6e−
2iπ 3
in Cartesian form.
NORMANHURST BOYS’ HIGH SCHOOL
√ Answer: −3 − 3i 3
26
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
Example 27
√ Evaluate the product (1 + i) 1 − i 3 in Cartesian form and polar form, to show √ π 1+ 3 that cos = √ . 12 2 2
Further exercises • Patel (2004, Ex 4C, Q1-10) • Arnold and Arnold (2000, Ex 2.2, Q1-4) • Lee (2006, Ex 2.6 Q8 onwards) NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
2.3.1
Properties
• Multiplication of z1 and z2 : moduli . . . . . . . . . . . . . . . . . . . , arguments . . . . . . . . , i.e. z1 z2 = r1 r2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) = r1 r2 ei(θ1 +θ2 ) Proof
Let z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 ) z1 z2 =
Proof
(via index laws and complex exponential) Let z1 = r1 eiθ1 and z2 = r2 eiθ2 . z1 z2 =
• Conjugates: if z = r (cos θ + i sin θ), then z = r (cos θ − i sin θ) = re−iθ Proof
Let z = r (cos θ + i sin θ).
Further exercises • Sadler and Ward (2014, Ex 1D) • Patel (2004, Ex 4C, Q11 onwards) • Lee (2006, Ex 2.5) NORMANHURST BOYS’ HIGH SCHOOL
27
28
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
• Powers:
Theorem 2 De Moivre’s Theorem (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) (Abraham De Moivre, 1667–1754. http://en.wikipedia.org/wiki/Abraham de Moivre)
Proof
(via complex exponential)
Proof
(by induction – for later in the course)
NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
29
Example 28 Simplify the following, expressing the answer in polar form: 3 (a) cos π3 + i sin π3 . 3π 3 (b) 2 cos 3π + i sin . 4 4 8 √ (c) 2 cos π6 − i sin π6 . 1 4 1 5 (d) cos 3π + i sin 3π ÷ 3 cos π8 − i sin π8 2 5 5
NORMANHURST BOYS’ HIGH SCHOOL
Answer: −1 π
Answer: 8ei 4 Answer: 16ei Answer:
2π 3
243 −i 39π e 40 16
30
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
Example 29 If |z1 | = 3, Arg(z1 ) = 2, |z2 | = 2 and Arg(z2 ) = 3, find the modulus and 2z1 2 9 argument of , Arg(z) = 2π − 5 . Answer: |z| = 20 5z2 3
Example 30 [1988 4U HSC Q4(a)] √ √ (a) Express z = 2 − i 2 in modulus-argument form. Answer: z = 2 cos
(b)
Hence write z 22 in the form a + ib, a, b ∈ R.
π 4
− i sin
π 4
Answer: z 22 = 222 i
NORMANHURST BOYS’ HIGH SCHOOL
Further arithmetic & algebra of complex numbers – Mod-arg & polar form
Example 31 (a)
(b)
√ If z1 = 1 + i and z2 = 3 − i, find the moduli and principal arguments of z1 , z1 √ z1 z2 and . Answer: z1 = 2 exp iπ , z2 = 2 exp − iπ , z = √1 exp 5iπ . 4 6 12 2 2 z2 1+i , find the smallest positive integer n such that z n is real, and If z = √ 3−i 1 evaluate z n for this integer n. Answer: n = 12, z 12 = − 64
Further exercises • Sadler and Ward (2014, Ex 7A) • Patel (2004, Ex 4D) • Lee (2006, Ex 2.9) • Arnold and Arnold (2000, Ex 2.2, Q6 onwards) NORMANHURST BOYS’ HIGH SCHOOL
31
Section 3 Locus problems
3.1 Summary Equation
Diagram
• |z| = r – Derivation of Cartesian equation:
– Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • |z − ω| = r – Derivation of Cartesian equation:
– Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
Locus problems – Summary
• |z − z1 | = |z − z2 |
– Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................... • Arg(z − z1 ) = α, α ∈ R
– Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................... • Arg(z − z1 ) − Arg(z − z2 ) = α, 0 < α < π – Origin: circle geometry theorem – Angle at the circumference subtended by the same arc/chord – Description: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................... • |z − z1 | + |z − z2 | = k, k ∈ R. – Description: . . . . . . . . . . . . . . . – Will return after Conic Sections topic is completed.
NORMANHURST BOYS’ HIGH SCHOOL
33
34
Locus problems – Examples
3.2 Examples Example 32 For the following: i. Describe the locus of z. ii. Draw a sketch. iii. Give the Cartesian equation of the locus. (a)
|z| = 2
(c)
|z + 2| = 1
(b)
zz = 16
(d)
|z + 2 + 3i| = 2
Answer: (a) x2 + y 2 = 4 (b) x2 + y 2 = 16 (c) (x + 2)2 + y 2 = 1 (d) (x + 2)2 + (y + 3)2 = 4
NORMANHURST BOYS’ HIGH SCHOOL
Locus problems – Examples
35
Example 33 Draw a sketch of locus of z if (a)
Arg(z) =
π 3
NORMANHURST BOYS’ HIGH SCHOOL
(b)
0 ≤ Arg(z) ≤
2π 3
(c)
Arg(z − 2 + 3i) =
π 4
36
Locus problems – Examples
Example 34 ! Draw a sketch of the locus of z if |z − 3| + |z + 3| = 12, and find its Cartesian
equation.
Answer:
x2 36
+
y2 27
=1
NORMANHURST BOYS’ HIGH SCHOOL
Locus problems – Examples
37
Example 35 Sketch the region in the Argand diagram defined simultaneously by 6 ≤ Re [(2 − 3i)z] < 12
and
Re(z) Im(z) > 0
Example 36 z is a complex number which simultaneously satisfies 2 ≤ |z + 3| ≤ 3
and
0 ≤ Arg(z + 3) ≤
π 3
Find the area and perimeter of the region in the Argand diagram determined by these restrictions on z. Answer: A = 5π units2 , P = 2 + 5π units 6 3
NORMANHURST BOYS’ HIGH SCHOOL
38
Locus problems – Examples
Example 37 Sketch the curve in the Argand diagram determined by Arg(z − 1) = Arg(z + 1) + π4 . Find its Cartesian equation. Answer: x2 + (y − 1)2 = 2, y > 0
Example 38 z satisfies |z − i| = Im(z) + 1. Sketch the locus of the point P representing z in the Argand diagram and write down its Cartesian equation. Answer: x2 = 4y
NORMANHURST BOYS’ HIGH SCHOOL
Locus problems – Examples
39
Example 39 Find the locus of z, given w is purely imaginary and w =
z−2 . z+1 Answer: Circle C
Further exercises • Sadler and Ward (2014, Ex 1F) • Patel (1990, Self Testing Ex 4.9, p.127) • Arnold and Arnold (2000, Ex. 2.5) • Fitzpatrick (1991, Ex 31(f)) • Lee (2006, Ex 2.7, 2.8) NORMANHURST BOYS’ HIGH SCHOOL
1 ,0 2
,r=
3 . 2
Section 4 Past HSC questions HSC problems that can be attempted without theory from other parts of the Extension 2 course.
4.1 2001 Extension 2 HSC Question 2 (a) (b)
(c)
Let z = 2 + 3i and w = 1 + i. Find zw and i. Express 1 +
1 in the form x + iy. w
√
3i in modulus-argument form. √ 10 in the form x + iy. ii. Hence evaluate 1 + 3i
2 2
Sketch the region in the complex plane where the inequalities |z + 1 − 2i| ≤ 3
and
−
2
3
π π ≤ arg z ≤ 3 4
both hold. (d)
Find all solutions of the equation z 4 = −1. Give your answers in modulus-argument form.
(e)
In the diagram the vertices of a triangle ABC are represented by the complex numbers z1 , z2 and z3 , respectively. The triangle is isosceles and right-angled at B. y D
3
b
A b
b
b
B O
C x
i. Explain why (z1 − z2 )2 = − (z3 − z2 )2 .
2
ii. Suppose D is the point such that ABCD is a square. Find the complex number, expressed in terms of z1 , z2 and z3 , that represents D.
1
40
Past HSC questions – 2001 Extension 2 HSC
41
Question 7 (a)
Suppose that z =
1 (cos θ + i sin θ) where θ is real. 2
i. Find |z|.
1
ii. Show that the imaginary part of the geometric series
3
1 + z + z2 + z3 + · · · = is
1 1−z
2 sin θ . 5 − 4 cos θ
iii. Find an expression for 1+
2
1 1 1 cos θ + 2 cos 2θ + 3 cos 3θ + · · · 2 2 2
in terms of cos θ. (b)
Consider the equation x3 − 3x − 1 = 0. p i. Let x = where p and q are integers having no common divisors other q than +1 and −1. Suppose that x is a root of ax3 − 3x + b = 0, where a and b are integers. Explain why p divides b and why q divides a. Deduce that x3 −3x−1 = 0 does not have a rational root. √ ii. Suppose that r, s and d are rational numbers and that d is irrational. √ Assume that r + s d is a root of x3 − 3x − 1 = 0. √ Show that 3r 2 s + s3 d − 3s = 0 and show that r − s d must also be a root of x3 − 3x − 1 = 0. 3 Deduce from this result and part (i), √ that no root of x − 3x − 1 = 0 can be expressed in the form r + s d with r, s and d rational. π iii. Show that one root of x3 − 3x − 1 = 0 is 2 cos . 9 You may assume the identity cos 3θ = 4 cos3 θ − 3 cos θ.
NORMANHURST BOYS’ HIGH SCHOOL
4
4
1
42
Past HSC questions – 2002 Extension 2 HSC
4.2 2002 Extension 2 HSC Question 2 (a)
(b)
Let z = 1 + 2i and w = 1 + i. Find, in the form x + iy, i. zw. 1 ii. . w
1 1
On an Argand diagram, shade in the region where the inequalities 0 ≤ Re(z) ≤ 2
and
3
|z − 1 + i| ≤ 2
both hold. (c)
It is given that 2 + i is a root of P (z) = z 3 + rz 2 + sz + 20
(d)
where r and s are real numbers. i. State why 2 − i is also a root of P (z).
1
ii. Factorise P (z) over the real numbers.
2
Prove by induction that, for all integers n ≥ 1,
3
(cos θ − i sin θ)n = cos (nθ) − i sin (nθ) (e)
Let z = 2 (cos θ + i sin θ). i. Find 1 − z.
1
ii. Show that the real part of
2
1 1 − 2 cos θ is 1−z 5 − 4 cos θ 1 iii. Express the imaginary part of in terms of θ. 1−z
1
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2003 Extension 2 HSC
43
4.3 2003 Extension 2 HSC Question 2 (a)
(b)
(c)
Let z = 2 + i and w = 1 − i. Find, in the form x + iy, i. zw. 4 ii. . z
1 1
Let α = −1 + i. i. Express α in modulus-argument form.
2
ii. Show that α is a root of the equation z 4 + 4 = 0.
1
iii. Hence, or otherwise, find a real quadratic factor of the polynomial z 4 +4.
2
Sketch the region in the complex plane where the inequalities |z − 1 − i| < 2
and
0 < arg(z − 1 − i) <
3
π 4
hold simultaneously. (d)
By applying De Moivre’s theorem and by also expanding (cos θ + i sin θ)5 , express cos 5θ as a polynomial in cos θ.
3
(e)
Suppose that the complex number z lies on the unit circle, and 0 ≤ arg(z) ≤ π2 .
2
Prove that 2 arg(z + 1) = arg(z).
NORMANHURST BOYS’ HIGH SCHOOL
44
Past HSC questions – 2004 Extension 2 HSC
4.4 2004 Extension 2 HSC Question 2 (a)
(b)
(c)
Let z = 1 + 2i and w = 3 − i. Find, in the form x + iy, i. zw. 10 ii. . z
1 1
√ Let α = 1 + i 3 and β = 1 + i. α i. Find in the form x + iy. β
1
ii. Express α in modulus-argument form.
2
iii. Given that β has the modulus-argument form √ π π β = 2 cos + i sin 4 4 α find the modulus-argument form of . β π iv. Hence find the exact value of sin . 12
1
1
Sketch the region in the complex plane where the inequalities |z + z| ≤ 1
and
3
|z − i| ≤ 1
hold simultaneously. (d)
The diagram shows two distinct points A and B that represent the complex numbers z and w respectively. The points A and B lie on the circle of radius r centred at O. The point C representing the complex number z + w also lies on this circle. C b
B b
A b
b
b
O
i. Using the fact that C lies on the circle, show geometrically that ∠AOB = 2π . 3
2
ii. Hence show that z 3 = w 3 .
2
iii. Show that z 2 + w 2 + zw = 0.
1 NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2005 Extension 2 HSC
45
Question 7 (b)
Let α be a real number and suppose that z is a complex number such that 1 = 2 cos α z i. By reducing the above equation to a quadratic equation in z, solve for z and use De Moivre’s theorem to show that z+
zn +
1 = 2 cos nα zn
1 ii. Let w = z + . Prove that z 3
2
w + w − 2w − 2 =
3
2
1 z+ z
1 1 2 3 + z + 2 + z + 3 z z
iii. Hence, or otherwise, find all solutions of
3
cos α + cos 2α + cos 3α = 0 in the range 0 ≤ α ≤ 2π.
4.5 2005 Extension 2 HSC Question 2 (a)
Let z = 3 + i and w = 1 − i. Find, in the form x + iy, i. 2z + iw. ii. zw. 6 iii. . w
(b)
(c)
1 1 1
√ Let β = 1 − i 3.
i. Express β in modulus-argument form.
2
ii. Express β 5 in modulus-argument form.
2
iii. Hence express β 5 in the form x + iy.
1
Sketch the region in the complex plane where the inequalities |z − z| < 2 hold simultaneously.
NORMANHURST BOYS’ HIGH SCHOOL
and
|z − 1| ≥ 1
3
46
(d)
Past HSC questions – 2005 Extension 2 HSC
Let ℓ be the line in the complex plane that passes through the origin and makes an angle α with the positive real axis, where 0 < α < π2 . ℓ b
Q
b
P
α b
O The point P represents the complex number z1 , where 0 < arg(z1 ) < α. The point P is reflected in the line ℓ to produce the point Q, which represents the complex number z2 . Hence |z1 | = |z2 |. i. Explain why arg(z1 ) + arg(z2 ) = 2α. ii. Deduce that z1 z2 = |z1 |2 (cos 2α + i sin 2α). π iii. Let α = and let R be the point that represents the complex number 4 zz.
2 1 1
1 2
Describe the locus of R as z1 varies. Question 6 (b)
Let n be an integer greater than 2. Suppose ω is an n-th root of unity and ω 6= 1. i. By expanding the left, show that 1 + 2ω + 3ω 2 + 4ω 3 + · · · + nω n−1 (ω − 1) = n ii. Using the identity
z −1 1 = , or otherwise, prove that z2 − 1 z − z −1
2
1
1 cos θ − i sin θ = cos 2θ + i sin 2θ − 1 2i sin θ provided that sin θ 6= 0.
2π 2π 1 + i sin , find the real part of . n n ω−1 4π 6π 8π 5 2π iv. Deduce that 1 + 2 cos + 3 cos + 4 cos + 5 cos =− . 5 5 5 5 2 v. By expressing the left hand side of the equation in part (iv) in terms of π 2π π cos and cos , find the exact value, in surd form, of cos . 5 5 5
iii. Hence, if ω = cos
1 1 3
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2006 Extension 2 HSC
47
4.6 2006 Extension 2 HSC Question 2 (a)
Let z = 3 + i and w = 2 − 5i. Find, in the form x + iy, i. z 2 . ii. zw. w iii. . z
(b)
i. Express
1 1 1
√
3 − i in modulus-argument form. √ 7 3 − i in modulus-argument form. ii. Express √ 7 iii. Hence express 3 − i in the form x + iy.
(c)
Find, in modulus-argument form, all solutions of z 3 = −1.
(d)
The equation |z − 1 − 3i| + |z − 9 − 3i| = 10 corresponds to an ellipse in the Argand diagram. i. Write down the complex number corresponding to the centre of the ellipse.
2 2 1 2
1
ii. Sketch the ellipse, and state the lengths of the major and minor axes.
3
iii. Write down the range of values of arg(z) for complex numbers z corresponding to points on the ellipse.
1
4.7 2007 Extension 2 HSC Question 2 (a)
Let z = 4 + i and w = z. Find, in the form x + iy, i. w. ii. w − z. z iii. . w
(b)
(c)
1 1 1
i. Write 1 + i in the form r (cos θ + i sin θ).
2
ii. Hence, or otherwise, find (1 + i)17 in the form a + ib, where a and b are integers.
3
The point P on the Argand diagram represents the complex number z, where z satisfies 1 1 + =1 z z Give a geometrical description of the locus of P as z varies.
3
NORMANHURST BOYS’ HIGH SCHOOL
48
(d)
Past HSC questions – 2007 Extension 2 HSC
The points P , Q and R on the Argand diagram represent the complex numbers z1 , z2 and a respectively. The triangles OP R and OQR are equilateral with unit sides, so |z1 | = |z2 | = |a| = 1. Let ω = cos π3 + i sin π3 . Im Q(z2 )
R(a) Re
P (z1 ) i. Explain why z2 = ωa.
1
ii. Show that z1 z2 = a2 .
1
iii. Show that z1 and z2 are the roots of z 2 − az + a2 = 0.
2
Question 8 (b)
i. Let n be a positive integer. Show that if z 2 6= 1, then n z − z −n 2 4 2n−2 1+z +z +···+z = z n−1 −1 z−z
2
ii. By substituting z = cos θ + i sin θ, where sin θ 6= 0 in to part (i), show that
3
1 + cos 2θ + · · · + cos(2n − 2)θ + i [sin 2θ + · · · + sin(2n − 2)θ] sin nθ = [cos(n − 1)θ + i sin(n − 1)θ] sin θ π iii. Suppose θ = . Using part (ii), show that 2n sin
3
π 2π (n − 1)π π + sin + · · · + sin = cot n n n 2n
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2008 Extension 2 HSC
49
4.8 2008 Extension 2 HSC Question 2 (a) (b)
(c)
Find real numbers a and b such that (1 + 2i)(1 − 3i)= a + ib.
2
√ 1+i 3 i. Write in the form x + iy, where x and y are real. 1+i √ ii. By expression both 1 + i 3 and 1 + i in modulus-argument form, write √ 1+i 3 in modulus-argument form. 1+i π iii. Hence find cos in surd form. 12 √ !12 1+i 3 iv. By using the result of part (ii), or otherwise, calculate . 1+i
The point P on the Argand diagram represents the complex number z = x + iy, which satisfies z2 + z2 = 8
2 3
1 1 3
Find the equation of the locus of P in terms of x and y. What type of curve is the locus? (d)
The point P on the Argand diagram represents the complex number z. The points Q and R represent the points ωz and ωz respectively, where ω = cos 2π + i sin 2π . The point M is the midpoint of QR. 3 3 Im P Q Re
O b
M R
S i. Find the complex number representing M in terms of z.
2
ii. The point S is chosen so that P QSR is a parallelogram.
2
Find the complex number represented by S.
NORMANHURST BOYS’ HIGH SCHOOL
50
Past HSC questions – 2009 Extension 2 HSC
4.9 2009 Extension 2 HSC Question 2 (a)
Write i9 in the form a + ib where a and b are real.
(b)
Write
(c)
The points P and Q on the Argand diagram represent the complex numbers z and w respectively.
1
−2 + 3i in the form a + ib where a and b are real. 2+i
1
Im b
P (z)
b
Q(w) Re
Copy the diagram into your writing booklet, and mark on it the following points: i. the point R representing iz
(d)
(e)
(f)
1
ii. the point S representing w
1
iii. The point T representing z + w.
1
Sketch the region in the complex plane where the inequalities |z − 1| ≤ 2 and − π4 ≤ arg(z − 1) ≤ π4 hold simultaneously.
2
i. Find all the 5th roots of −1 in modulus-argument form.
2
ii. Sketch the 5th roots of −1 on an Argand diagram.
1
i. Find the square roots of 3 + 4i.
3
ii. Hence, or otherwise, solve the equation
2
z 2 + iz − 1 − i = 0
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2010 Extension 2 HSC
51
4.10 2010 Extension 2 HSC Question 2 (a)
Let z = 5 − i. i. Find z 2 in the form x + iy.
1
ii. Find z + 2z in the form x + iy.
1
i in the form x + iy. z √ i. Express − 3 − i in modulus-argument form. √ 6 ii. Show that − 3 − i is a real number.
iii. Find (b)
2 2 2
(c)
Sketch the region in the complex plane where the inequalities 1 ≤ |z| ≤ 2 and 0 ≤ z + z ≤ 3 hold simultaneously.
(d)
Let z = cos θ + i sin θ where 0 < θ <
2
π . 2
On the Argand diagram the point A represents z, the point B represents z 2 and the point C represents z + z 2 . Im b
b
C
B b
A
θ b
Re
O Copy or trace the diagram into your writing booklet. i. Explain why the parallelogram OACB is a rhombus. 3θ 2 θ iii. Show that |z + z 2 | = 2 cos . 2 ii. Show that arg (z + z 2 ) =
1 2
iv. By considering the real part of z + z 2 , or otherwise, deduce that cos θ + cos 2θ = 2 cos
4.11 2011 Extension 2 HSC See Examples 21 on page 21 and 18 on page 18. NORMANHURST BOYS’ HIGH SCHOOL
1
θ 3θ cos 2 2
1
52
Past HSC questions – 2012 Extension 2 HSC
4.12 2012 Extension 2 HSC 1.
1
Let z = 5 − i and w = 2 + 3i. What is the value of 2z + w? (A) 12 + i
2.
(B) 12 + 2i
(C) 12 − 4i
(D) 12 − 5i
The complex number z is shown on the Argand diagram below. y
b
1
z x
O
Which of the following best represents iz? (A)
(C)
y b
y
iz
x
O
x
O
b
iz
(B)
(D)
y iz
y
b
O
x
x
O
b
iz
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2012 Extension 2 HSC
53
Question 11 (a)
√ 2 5+i in the form x + iy, where x and y are real. Express √ 5−1
2
(b)
Sketch the region in the complex plane where the inequalities
2
|z + 2| ≥ 2
and
|z − i| ≤ 1
both hold. (d)
i. Write z =
√
3 − i in modulus-argument form.
2
ii. Hence express z 9 in the form x + iy, where x and y are real.
1
Question 12 (d)
On the Argand diagram the points A1 and A2 correspond to the distinct complex numbers u1 and u2 respectively. Let P be a point corresponding to a third complex number z. Points B1 and B2 are positioned so that △A1 P B1 and △A2 B2 P , labelled in an anti-clockwise direction, are right-angled and isosceles with right angles at A1 and A2 respectively. The complex numbers w1 and w2 correspond to B1 and B2 respectively. y b
B1 (w1 )
B2 (w2 )
b
P (z) b
A1 (u1) b
A2 x
i. Explain why w1 = u1 + i(z − u1 ).
1
ii. Find the locus of the midpoint of B1 B2 as P varies.
2
NORMANHURST BOYS’ HIGH SCHOOL
54
Past HSC questions – 2013 Extension 2 HSC
4.13 2013 Extension 2 HSC 5.
Which region on the Argand diagram is defined by (A)
π π ≤ |z − 1| ≤ ? 4 3
(C)
y π 3
π 4 bc
y
x
π 4 bc
π 3
(B)
(D)
y
b
1−
π 3
b
1−
π 4 b
1+
1
π 4
b
1+
π 3
x
1
1
y
x
b
1−
π 3
b
1−
π 4 b
1+
1
π 4
b
1+
π 3
x
Question 11 (a)
√ √ Let z = 2 − i 3 and w = 1 = i 3. i. Find z + w.
1
ii. Express w in modulus-argument form.
2
iii. Write w 24 in its simplest form. (c)
Factorise z 2 + 4iz + 5.
2
(e)
Sketch the region on the Argand diagram defined by z 2 + z 2 ≤ 8.
2
Question 14 (b)
!
Let z2 = 1 + i and, for n > 2, let zn = zn−1 1 +
3 i |zn−1 |
Use mathematical induction to prove that |zn | =
√
n for all integers n ≥ 2.
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2013 Extension 2 HSC
55
Question 15 (a)
The Argand diagram shows complex numbers w and z with arguments φ and θ respectively, where φ < θ. The area of the triangle formed by O, w and z is A. Im b
z
b
θ
Show that zw − wz = 4iA
NORMANHURST BOYS’ HIGH SCHOOL
φ
w Re
3
56
Past HSC questions – 2014 Extension 2 HSC
4.14 2014 Extension 2 HSC 4.
8.
π π Given z = 2 cos + i sin , which expression is equal to (z)−1 ? 3 3 1 π π 1 π π (C) (A) cos − i sin cos + i sin 2 3 3 2 3 3 π π π π (B) 2 cos − i sin (D) 2 cos + i sin 3 3 3 3 The Argand diagram shows the complex numbers w, z and u, where w lies in the first quadrant, z lies in the second quadrant and u lies on the negative real axis. Im z w b
1
b
Re
b
u
1
O
Which statement could be true? (A) u = zw and u = z + w
(C) z = uw and u = z + w
(B) u = zw and u = z − w
(D) z = uw and u = z − w
Question 11 (a)
(c)
Consider the complex numbers z = −2 − 2i and w = 3 + i. i. Express z + w in modulus-argument form. z ii. Express in the form x + iy, where x and y are real numbers. w Sketch the region in the Argand diagram where |z| ≤ |z − 2| and π π − ≤ arg z ≤ . 4 4
2 2 3
NORMANHURST BOYS’ HIGH SCHOOL
Past HSC questions – 2015 Extension 2 HSC
57
4.15 2015 Extension 2 HSC 2.
5.
What value of z satisfies z 2 = 7 − 24i?
1
(A) 4 − 3i
(C) 3 − 4i
(B) −4 − 3i
(D) −3 − 4i
Given that z = 1 − i, which expression is equal to z 3 ? √ 3π 3π (A) z = 2 cos − + i sin − 4 4
1
√
3π 3π (B) z = 2 2 cos − + i sin − 4 4 (C) z =
√
3π 3π 2 cos + i sin 4 4
√
3π 3π (D) z = 2 2 cos + i sin 4 4 9.
1
The complex number z satisfies |z − 1| = 1.
What is the greatest distance that z can be from the point i on the Argand diagram? √ √ √ (C) 2 2 (D) 2 + 1 (A) 1 (B) 5 Question 11 (a) (b)
4 + 3i in the form x + iy, where x and y are real. 2−i √ π π Consider the complex numbers z = − 3 + i and w = 3 cos + i sin . 7 7 i. Evaluate |z|. Express
ii. Evaluate arg(z). iii. Find the argument of
2
1 1
z . w
1
Question 12 (a)
The complex number z is such that |z| = 2 and arg(z) =
π . 4
Plot each of the following complex numbers on the same half-page Argand diagram. i. z.
1
ii. u = z 2 .
1
iii. v = z 2 − z.
1
NORMANHURST BOYS’ HIGH SCHOOL
References Arnold, D., & Arnold, G. (2000). Cambridge Mathematics 4 Unit (2nd ed.). Cambridge University Press. Fitzpatrick, J. B. (1991). New Senior Mathematics – Four Unit Course for Years 12. Rigby Heinemann. Lee, T. (2006). Advanced Mathematics: A complete HSC Mathematics Extension 2 Course (2nd ed.). Terry Lee Enterprise. Patel, S. K. (1990). Excel 4 Unit Maths. Pascal Press. Patel, S. K. (2004). Maths Extension 2 (2nd ed.). Pascal Press. Sadler, D., & Ward, D. (2014). SGS Mathematics Year 12 Extension 2 Notes and Exercises. Sydney Grammar School.
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