4-Chapter Complex Number

4 Complex Numbers INTRODUCTION Ø Consider a simple quadratic equation x2 + 4 = 0. Clearly there is no solution of this equation in the set of real numbers. To permit the solution of such equations the set of complex numbers is introduced. Ø The solutions of the above equation are given by x2 = –4 Ø

Þ x = ± - 4 = ±2 – 1 = ±2 i. Swiss Mathematician Euler introduced the symbol i (iota)

for positive square root of – 1, i.e., i =

Ø

Ø

Ø

-1 .

The square root of a negative number is called an imaginary number. Now for any two real numbers x and y, we can form a new number x + iy. This number x + iy is called a complex number. The set C of complex numbers is therefore defined by C = {x + iy | x Î R, y Î R} The extension of concept of numbers from real numbers to complex numbers enabled us to solve any polynomial equation. A complex number is deonoted by a single letter such as z, w etc. Given a complex number z = x + iy, x is called its Real part and y its Imaginary Part and we denote x = Re(z) and y = Im (z) If y = 0, then z = x is a purely real number. If x = 0, then z = iy is a purely imaginary number. The complex number 0 = 0 + i0 is both purely real as well as purely imaginary. Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are said to be equal if and only if x1 = x2 and y1 = y2.

AXIOMATIC APPROACH TOWARDS THE COMPLEX NUMBER SYSTEM Ø A complex number is defined as an ordered pair (x, y) of real numbers x and y. Thus, C = {(x, y) | x Î R, y Î R}

If z = (x, y) then x = Re(z) and y = Im (z), z = (x, y) is equivalent to z = x + iy. Thus, i is equivalent to (0, 1) and z = (x, –y). IMPORTANT RESULTS TO BE MEMORISED ABOUT i i (greek letter iota) represents positive square root of –1, so, i = - 1 . It is called imaginary unit. We have 1. i 2 = -1, i 3 = -i, i 4 = 1, i 5 = i, ........etc. Thus for any integer k, i4k = 1, i4k+1 = i, i4k+2 = –1, i4k+3 = –i. That is if power of i is m, m Î N , then divide m by 4 and find the remainder. If the remainder is zero, then im = 1 If the remainder is one, then im = i If the remainder is two, then im = –1 If the remainder is three, then im = –i 2. The sum of four consecutive powers of i is zero, for example, i12 + i13 + i14 + i15 = 0

a ´ b = ab is true only if at least one number is non negative or zero.

3. For any two real numbers a and b, If both a and b are negative then

a ´ b ¹ ab

In fact if a > 0 and b > 0 then -a ´ -b = -1 a ´ -1 b = i a ´ i b = - ab OPERATIONS ON COMPLEX NUMBERS Suppose that z1 = (x1, y1) and z2 = (x2, y2) be two complex numbers, that is, z1 = x1 + iy1 and z2 = x2 + iy2 (i) Equality : z1 = z2, if x1 = x2 and y1 = y2 (ii) Addition : z1 + z2 = (x1 + x2, y1 + y2) or equivalently z1 + z2 = (x1 + x2) + i (y1 + y2)

2

Ph y si cs

(iii) Subtraction : z1 – z2 = (x1 – x2, y1 – y2) or equivalently z1 – z2 = (x1 – x2) + i (y1 – y2) (iv) Multiplication : z 1z 2 = (x 1x2 – y1y2, x1y2 + x2y1) or equivalently z1z2 = (x1x2 – y1y2) + i (x1y2 + x2y1) (v) Division : If z2 ¹ 0 then

Clearly, modulus of a complex number is a real number. Again let z = x + iy be a complex number. Then the complex number x – iy is called the conjugate of z and is denoted by z or z*. Thus, we have Re ( z ) = Re (z) and Im ( z ) = – Im (z). Note : 1. The additive inverse of ‘z’ is ‘– x – iy’ and conjugate of ‘z’ is ‘x – iy’. 2. The multiplicative inverse of a non-zero complex number ‘z’ can be given by

z1 æç x1 x 2 + y1 y 2 y1 x 2 - y 2 x1 ö÷ = , z 2 çè x 22 + y 22 x 22 + y 22 ÷ø

or equivalently

z1 æç x1x 2 + y1 y 2 = z 2 çè x 22 + y 22

ö ÷+ ÷ ø

æ y x - y 2 x1 ö ÷ iç 1 2 ç x 2 + y2 ÷ 2 2 è ø

z -1 =

(vi) Multiplication by a real number : If z = (x, y) and m Î R then mz = (mx, my) or equivalently if z = x + iy then mz = mx + imy ALGEBRA OF OPERATIONS If z 1, z 2 and z3 belong to set C of complex numbers, then following properties hold. z (1) Closure Property : z1 ± z2; z1z2 and 1 , z2 ¹ 0 all also z2

belong to C. (2) Commutative Property : z1 + z2 = z2 + z1 and z1z2 = z2z1 (3) Associative Property : z1 + (z2 + z3) = (z1 + z2) + z3 and z1 (z2z3) = (z1z2)z3 (4) Cancellation Property : z1 + z3 = z2 + z3 Þ z1 = z2 and z1z3 = z2z3 Þ z3 = 0 or z1 = z2 (5) Distributive Property : z1 (z2 + z3) = z1z2 + z1z3 (6) Existence of Identity : 0 = (0, 0) is additive identity, i.e. 0 +z =z+0 =z " zÎC 1 = (1, 0) is multiplicative identity, i.e. 1(z) = (z)1 = z " z Î C (7) Existence of Inverse : For every complex number z = (x, y), we may get a unique number–z = (–x, –y) such that z + (–z) = (–z) + z = 0. (–z) is Additive Inverse. For every complex number z = (x, y), z ¹ 0 we may get a unique number z–1 or

x –y 1 æç , =ç 2 2 2 2 z èx +y x +y

ö ÷ ÷ such that ø

|z|=

a + b = {Re ( z )} + {Im ( z )} 2

2

2

2

z z = | z |2

or

IMPORTANT RESULTS TO BE MEMORISED ABOUT CONJUGATE (i)

(z ) = z

(ii) z1 ± z 2 = z1 ± z 2

(iii) z1z 2 = z1z 2 Þ (z n ) = ( z ) n , n Î N æ z1 ö z1 , z2 ¹ 0 (v) çç ÷÷ = è z 2 ø z2

(iv) (i z) = -i z

(vi) z + z = 2 Re( z) , which is a purely real number (vii) z - z = 2 i Im(z ) , which is a purely imaginary number.. (viii) z = z if and only if z is purely real. (ix) z = - z if and only if z is purely imaginary.. (x) If f(z) is a polynomial in a complex variable z, then

f (z) = f (z) [where f means the complex coefficients are replaced by their conjugate is] (xi) z1z 2 + z1z 2 = 2 Re( z1z 2 ) = 2 Re( z1z 2 ) a1

a2

(xii) If z = b1 c1

b2 c2

a3

a1 b3 then z = b1 c3 c1

a2 b2

a3 b3

c2

c3

where ai, bi, ci (i = 1, 2, 3) are complex numbers.

æ1ö æ1ö æ1ö zç ÷ = ç ÷z = 1 . ç ÷ is multiplicative inverse. z z è ø è ø èzø

[NOTE : A set with two operations on it satisfying all above properties is called a Field.] (8) The order relations 'greater than' and 'less than' are not defined for non real complex numbers. The inequalities like –2i < 0; 1 + 2i > 1; i – 1< i are meaning less. The Modulus and the conjugate of a Complex Number Let z = x + iy be a complex number. Then the modulus (absolute value) of z, denoted by | z | is defined as follows :

1 x - iy z = = 2 2 x + iy x + y | z |2

ILLUSTRATIVE EXAMPLES 4n +7

1.

The value of the sum

å ik

is

k =1

(a) 0 (c) –1

(b) 1 (d) i or –i, depending on n is even or odd 4n +7

Sol. We have,

å

k =1

4n + 7

i k = i + i 2 + i3 +

å i k = i -1 - i + 0 = -1

k =4

[Note that, k = 4 to 4n + 7 contains 4n + 4 terms, a multiple of 4] Answer (c)

Laws of Motion 2.

(1 + i )x - 2i (2 - 3i ) y + i + = i, x , y Î R , then 3+i 3-i (a) x = 3, y = –1 (b) x = –3, y = 1 (c) x = 3, y = 1 (d) x = –3, y = –1

Then z1 Im( z1 z2 ) = ( x1 + iy1 ) Im éë( x2 – iy2 )( x3 + iy3 ) ùû

(1 + i )x - 2i (2 - 3i ) y + i + =i 3+i 3-i Multiplying both sides by (3 + i) (3 – i), we get [(1 + i)x – 2i] (3 – i) + [(2 – 3i)y + i] (3 + i) = (3 + i) (3 – i)i Þ (4x + 9y – 3) + (2x – 7y – 3)i = 10i Þ 4x + 9y – 3 = 0 and 2x – 7y – 13 = 0 Solving these equations, we get x = 3, y = –1 Answer (a)

= x1 ( x2 y3 – x3 y2 ) + iy1 ( x2 y3 – x3 y2 )

If

= ( x1 + iy1 ) Im éë( x2 x3 + y2 y3 ) + i ( x2 y3 – x3 y2 ) ùû = ( x1 + iy1 )( x2 y3 – x3 y2 )

Sol. We have

3.

(cos x + i sin x)(cos y + i sin y) If = A + iB , then (cot u + i )(1 + i tan v) (a) A = sinu cosv cos (x + y – u – v) (b) B = sinu cosv sin(x + y – u – v) (c) A = cosu sinv cos(x + y – u – v) (d) B = cosu sinv sin(x + y – u – v)

Sol. We have, =

Similarly, z 2 Im( z 3z 2 ) = x 2 ( x 3 y1 – x1y3 ) + iy 2 ( x 3 y1 – x1 y3 ) and z3 Im(z 1 z 2 ) = x 3 ( x1y 2 – x 2 y1 ) + iy 3 ( x1y 2 – x 2 y1 )

5.

Clearly z1 Im( z 2 z 3 ) + z 2 Im( z 3 z1 ) + z 3 Im( z 1 z 2 ) = 0 Answer (d) If (x + iy)1/3 = a + ib, where x, y, a, b Î R, show that

x y - = -2( a 2 + b 2 ) a b Sol. (x + iy)1/3 = a + ib Þ x + iy = (a + ib)3 i.e., x + iy = a3 + i3b3 + 3iab (a + ib) = a3 – ib3 + i 3a2b – 3ab2 = a3 – 3ab2 + i (3a2b – b3) Þ x = a3 – 3ab2 and y = 3a2b – b3

Thus

(cos x + i sin x )(cos y + i sin y ) sin v ö æ cos u öæ + i ÷ç1 + i ç ÷ sin u cos vø øè è

x y = a2 – 3b2 and = 3a2 – b2 a b

x y - = a2 – 3b2 – 3a2 + b2 a b = – 2a2 – 2b2 = – 2(a2 + b2). 6. Solve the equation z2 = z , where z = x + iy Sol. z2 = z Þ x2 – y2 + i2xy = x – iy Therefore, x2 – y2 = x ...... (1) and 2xy = – y ...... (2)

So,

=

sin u cos v(cos x + i sin x )(cos y + i sin y) (cos u + i sin u )(cos v + i sin v)

=

sin u cos v[cos( x + y) + i sin( x + y)] [cos( u + v) - i sin( u + v )] ´ [cos( u + v) + i sin( u + v)] [cos( u + v) - i sin( u + v )]

=

4.

(cos x + i sin x)(cos y + i sin y) (cot u + i )(1 + i tan v)

3

sin u cos v[cos( x + y - u - v) + i sin( x + y - u - v)]

From (2), we have y = 0 or x = –

cos 2 ( u + v ) + sin 2 ( u + v)

1 2

= sin u cos v cos( x + y - u - v) +i sin u cos v sin( x + y - u - v)

When y = 0, from (1), we get x2 – x = 0, i.e., x = 0 or x = 1.

\ A = sin u cos v cos( x + y - u - v)

When x = –

and B = sin u cos v sin(x + y - u - v) Answers (a, b) If z1z2z3 are there complex numbers then the value of

1 , from (1), we get 2

1 1 3 3 + or y2 = , i.e., y = ± . 4 2 4 2 Hence, the solutions of the given equation are

y2 =

z1 Im ( z 2 z 3 ) + z 2 Im(z3z1 ) + z 3 Im(z 1 z 2 ) is (a) Re (z1z2z3) (b) Im (z1z2z3) (c) Re (z1 + z2 + z3) (d) 0

0 + i0, 1 + i0, –

Sol. Let z1 = x1 + iy1, z2 = x2 + iy2 , z3 = x3 + iy3

1 3 1 3 +i ,– –i 2 2 2 2

4.1 Solve following problems with the help of above text and examples. 1. 2.

+i

588

+i

+i

584

i +i +i (a) 2 (b) –2 1 + i2 + i4 ... + i2n is (a) Positive (c) 0

578

+i +i (c) 1

574

The value of

i

592 582

+i

590 580

586 576

13

3.

(a) i

- 1 is

(d) –1

(b) Negative (d) Can’t be determined

The value of

å (i n + i n+1) , where i =

n =1

(b) i – 1

- 1 equals

(c) – i

(d) 0 n

4.

æ 2i ö ÷ is a positive The least positive integer n such that ç è1+ i ø integer is (a) 2 (b) 4 (c) 8 (d) 16

4 5.

Ph y si cs The multiplicative inverse of (6 + 5i ) 2 is 11 60 - i 61 61 9 60 - i (c) 61 61

(a)

6.

7.

(b)

11 60 + i 61 61

(d) None of these

æ 2 ö æ 11 ö (a) ç ÷ + ç ÷i è 25 ø è 25 ø

æ 2 ö æ 11 ö (b) ç ÷ - ç ÷i è 25 ø è 25 ø

æ 2 ö æ 11 ö (c) ç - ÷ + ç ÷i è 25 ø è 25 ø

æ 2 ö æ 11 ö (d) ç - ÷ - ç ÷i è 25 ø è 25 ø

2. (d)

(c) -

3. (b)

4. (c)

5. (a)

6. (c)

9.

O

Ø

Ø

1 9 - i 4 4

1 is 1 - cos q + 2i sin q

(b)

q 2 tan 1 2 +i 5 + 3 cos q 5 + 3 cos q

(d) None of these

a - ib = x - iy, then 3 a + ib =

ANSWER KEY 7. (d) 8. (b)

9. (c)

(b) x - iy

(c) y + ix

(d) y - ix

10. (a)

IMPORTANT RESULTS ABOUT MODULUS 1. | z |³ 0 , | z |= 0 Û z = 0 . 2. zz =| z |2 ( majority of the complex equations are solved using this property) 3. | z |=| z |=| -z |=| -z | 4. | z1z 2 |=| z1 || z 2 |Þ| z n |=| z |n

x

z1 | z1 | 5. z = | z | , z 2 ¹ 0 2 2

Every complex number x+iy can be represen ted geometrically as a unique point P (x,y) in the xoy plane with x-coordinate representing its real part and y-coordinate representing its imaginary part. The Point (x, 0) on the x-axis represents the purely real number x. As such x-axis is called the real axis. Similarly, the point (0, y) on the y-axis represents purely imaginary number iy. Therefore, y-axis is called the imaginary axis. The plane having a complex number assigned to each of its points, is called the complex plane or Argand plane or Guassian plane. This representation of complex numbers as points in the plane is known as Argand diagram. The distance from the origin to the point P(x, y) is defined as the MODULUS (or absolute value) of the complex number z = x + iy, denoted by | z |, thus | z | =

Ø

3

1 9 + i 4 4

(d) -

(a) x + iy

Q (x,–y)

Ø

The value of

10. If

P (x,y)

M (x,o)

1 9 + i 4 4

q 2 cot 1 2 -i (c) 5 + 3 cos q 5 + 3 cos q

y

N (0, y)

(b)

q 2 tan 1 2 (a) 5 + 3 cos q - i 5 + 3 cos q

GRAPHICAL REPRESENTATION OF COMPLEX NUMBERS

Ø

1 9 - i 4 4

(a)

3 + 4i The multiplicative inverse of is 4 - 5i 8 31 8 31 + i - i (b) (a) 25 25 25 25 8 31 - i (d) None of these (c) 25 25 2-i is The conjugate complex number of (1 - 2i) 2

1. (b)

3 öæ 3 + 4i ö æ 1 + ÷ç ÷ is The value of ç 1 2 i 1 + i øè 2 - 4i ø è

8.

x 2 + y2

The conjugate z of complex number z is represented by the point Q, which is the mirror image of P on the x–axis.

z z =1Þ , is a unimodular complex number |z| |z|

6.

( z ¹ 0 ). 7. | z1 + z 2 |2 = | z1 | 2 + | z 2 | 2 +2 Re( z1z 2 ) 8. | z1 - z 2 |2 = | z1 |2 + | z 2 |2 -2 Re( z1 z 2 ) 9. | z1 + z 2 |2 + | z1 - z 2 |2 = 2(| z1 |2 + | z 2 |2 ) 10. | az1 + bz 2 |2 + | bz1 - az 2 |2 = (a 2 + b 2 )(| z1 |2 + | z 2 |2 ),

where a, b Î R 11. If z1, z 2 ¹ 0 then | z1 + z 2 |2 =| z1 |2 + | z 2 | 2 Û purely imaginary.

z1 is z2

5

Laws of Motion POLAR FORM (OR TRIGONOMETRICAL FORM) OF COMPLEX NUMBERS y Let P r epresents the n onzero complex number z = x+iy. Let the directed line regment OP be of length r and makes an angle q with the positive direction of the x–axis (q in radians)

P (z) r=

For example, if z = 1 + i , then arg(z) =

|z|

y

p 4

Case II : If x < 0, y > 0, then the point P lis in the second

q O

Following cases may arise Case I : If x > 0, y > 0, then the point P lies in the first quadrant and then q = arg z = a

quadrant and then q = arg z = p - a X

M

X

The point P is uniquely determined by the ordered pair of real numbers (r, q) called the polar coordinates of the point P. Clearly, x = r cos q, y = r sin q,

For example, if z = -1 + 3i , then tan a =

y x Thus z=r(cosq + isin q) is the poar from of z. r is the modulus of the number z and q is called th e ARGUMENT (or AMPLITUDE) of the number z, denoted by arg (z) or amp (z) Hence.

so arg( z ) = p -

r= x 2 + y 2 , tanq =

r =| z |= x 2 + y 2 and q = arg (z) = tan -1 y x z = r(cosq + i sinq) is also written as r cis(q) Ø Note that q is not defined uniquely, In fact q is the solution of simultaneous equations

cos q =

x x 2 + y2

y

and sin q =

x 2 + y2

If z = 1 + i, r = x + y = 1 + 1 = 2 ; 2

2

2

2

p 2p = 3 3

y P

x¢

q

a

O

Case III : If x < 0, y < 0 then the point P lies in the third y quadrant and then q = arg z = -(p - a ) = a - p O x¢ For example, if z = -1 - i a -1 =1 -1

p p p , + 2p,..........., + 2kp, k Î I 4 4 4 Any two arguments of a complex number differ by a number which is a multiple of 2 p. The unique value of q, such that -p < q £ p is called the principal value of the Argument.

For example, if z = 3 - i, then

P(z)

tan a =

y , x ¹ 0, y ¹ 0 x

x¢ O y¢

x

P

y¢

3p p so arg(z) = -(p - ) = 4 4 Case IV : If x > 0, y < 0 then the point P lies in the fourth quadrant and then q = arg z = -a y

tan a =

-1

1

x¢

O

a

x

= q 3 3 P p so, arg( z ) = y¢ 6 Case V : If y = 0, then z is purely real and P lies on real axis, and z = x, so arg (z) = 0 if x > 0; arg (z) = p if x < 0 æ 1ö For example arg(3) = 0 and argç - ÷ = p è 2ø Case VI : If x = 0, then z = iy is a purely imaginary number and P lies on imaginary axis

y a=q x

x

q

then tan a =

y

x

y¢

y p tan q = = 1 Þ q = x 4 p pö æ Polar from of z = 2 ç cos + i sin ÷ 4 4ø è Clearly the possible arguments of the number z=1+i are the following angles :

WORKING RULE FOR FINDING PRINCIPAL ARGUMENT Let z = x + iy has image P on the argand plane and

3 = 3 -1

So, arg (z) =

p p if y > 0 and arg (z) = - if y < 0 2 2

For example, arg( 2 i) =

p p and arg(-100 i ) = 2 2

6

Ph y si cs IMPORTANT RESULTS ABOUT ARGUMENT

1.

arg(z ) = - arg(z )

2.

arg(z1z 2 ) = arg(z1 ) + arg(z 2 ) + 2kp

3.

æz ö argçç 1 ÷÷ = arg(z1 ) - arg(z 2 ) + 2kp è z2 ø

4.

æzö arg ç ÷ = 2arg(z) + 2kp èzø

5.

arg(z n ) = n arg(z) + 2kp

7.

If z1 and z2 are any two complex numbers then | z1 + z12 - z 2 2 | + | z1 - z12 - z 22 | is equal to (a) | z1 | (b) | z 2 |

(c) | z1 + z 2 | (d) | z1 + z 2 | + | z1 - z 2 | Sol. (Trick ) The nature of the problem suggests at once that we shold use the formula | z1 + z 2 |2 + | z1 - z 2 |2 = 2 (| z1 |2 + | z 2 | 2 )

We have (| z1 + z12 - z 22 | + | z1 - z12 - z 22 ) 2 =| z1 + z12 - z 22 |2 + | z1 - z12 - z 22 | 2 +2 | z12 - z12 + z 22 |

Where k = 0, –1 or 1 would be taken so that the value comes out into the principal value region

= 2éê| z1 | 2 + | z12 - z 22 |2 ùú + 2 | z 2 |2 ë û

6.

æ z2 ö æ z1 ö If argçç ÷÷ = q the argçç ÷÷ = 2kp - q , k Î I è z1 ø è z2 ø

7.

arg(z ) - arg(-z ) = ± p

= | z1 + z 2 |2 + | z1 - z 2 |2 +2 | z1 + z 2 || z1 - z 2 |

p + arg(z ) 2 | z1 + z 2 |=| z1 - z 2 |Û arg(z1 ) - arg(z 2 ) = p / 2

= (| z1 + z 2 | + | z1 - z 2 |) 2

8. 9.

2 2 2 2 = 2(| z1 | + | z 2 | ) + 2 | z1 - z 2 |

arg(iz ) =

\ | z1 + z12 - z 22 | + | z1 - z12 - z 22 | =| z1 + z 2 | + | z1 - z 2 |

10. | z1 + z 2 |=| z1 | + | z 2 |Û arg(z1 ) = arg(z 2 )

n

8. EULER'S NOTATION It can be shown that e iq = cos q + i sin q, e -iq = cos q - i sin q \ e z = e x + iy = e x .e iy = e x (cos y + i sin y ) Also r(cos q + i sin q) = reiq Again, cos q =

Sol.

LOGARITHM OF A COMPLEX NUMBER

1+ i ö The smallest positive integer n for which æç ÷ = 1 is 1 è –iø (a) 4 (b) 3 (c) 2 (d) 1

Sol. As a + ib = 1 Þ a 2 + b 2 = 1 (1 + b + ai) 2

a + b + ai

= 1 + b – ai = (1 + b – ai)(1 + b + ai) =

Then log e (z) = log e (reiq ) = log e r + iq

=

As such the argument of a complex number is not unique, the log of a complex number cannot be unique. In general, log e (z) = log e | z | +i[2kp + arg(z )], k Î I pö æ ip / 2 = iç 2kp + ÷, k Î I So, i p is For example, log( i ) = log e 2ø è 2

one of the values of log (i) p p æpö æpö Also, log(log i ) = log(i ) = log i + log ç ÷ = i + log ç ÷ 2 2 è2ø è2ø

[Taking principal value only]

æ 1+ i ö n \ç ÷ =1Þ i =1 è1– i ø

Answer (a) 1 + b + ai If a + ib = 1. the simplified form of is 1 + b - ai (a) b + ai (b) a + bi (c) (1 + b)2 + a2 (d) ai

Let z = x + iy = r(cos q + i sin q) = reiq

So, log e (z ) = log e | z | +i arg(z)

n

1 + i 1 + i 1 + i 1 + i 2 + 2i 2i = ´ = = =i 1– i 1– i 1+ i 2 1 – i2

Clearly the smallest value of n is 4.

9.

e iq + e -iq e iq - e -iq and sin q = 2 2i

Answer (d)

(1 – a 2 ) + b 2 + 2b + 2ai + 2abi 1 + (a + b ) + 2b 2

2

=

(1 + b) 2 – a 2 + 2(1 + b)ai (1 + b) 2 + a 2

b 2 + b 2 + 2b + 2ai + 2abi 1 + 1 + 2b

b 2 + b + ai + abi b(1 + b ) + a (1 + b)i = = b + ai 1+ b 1+ b Answer (a) 10. The argument of the complex number 1 + sin a - i cos a is a p a a p a p + (a) (b) (c) (d) + 2 4 2 2 4 2 2 Sol. Let 1+ sina – icosa = r(cosq + isinq) Then r cos q = 1 + sina and r sinq = – cosa

=

æ

r2 = (1 + sina)2 + (–cosa)2 = 2 + 2 sina = 2ç cos è

r=

a aö æ æp aö 2 ç cos + sin ÷ or 2 cos ç - ÷ 2 2ø è è4 2ø

a aö + sin ÷ 2 2ø

2

Laws of Motion - cos a = Also, tan q = 1 + sin a

æp ö - sinç - a ÷ è2 ø [Note the step] æp ö 1 + cos ç - a ÷ è2 ø

Y P(-4, 4 3) q

æp aö æp aö - 2 sin ç - ÷ cos ç - ÷ è 4 2 ø è 4 2 ø = - tan æ p - a ö = tan æ a - p ö ç ÷ ç ÷ = æp aö è4 2ø è 2 4ø 2 cos 2 ç - ÷ è4 2ø

\q=

a p 2 4

Convert the complex number

Sol. The given complex number

=

-16 (1 - i 3) 1- (i 3)

2

X'

X

O

Y'

æp aö Hence, Modulus = 2 cosç - ÷ and argument è4 2ø a p = - . 2 4 Answer (a)

11.

7

=

-16

-16

Hence, cos q = –

into polar form.

1+ i 3

=

1+ i 3

Let – 4 = r cos q, 4 3 = r sin q By squaring and adding, we get 16 + 48 = r2 (cos2 q + sin2 q) which gives r2 = 64, i.e., r = 8

-16 1+ i 3

×

1- i 3 1- i 3

-16 (1 - i 3) = – 4 (1 - i 3) 1+ 3

= – 4 + i4 3

q=p–

3 1 , sin q = 2 2

2p p = 3 3

2p 2p ö æ + i sin Thus, the required polar form is 8 ç cos ÷ 3 3 ø è 12. If | z2 – 1| = | z |2 + 1, then show that z lies on imaginary axis. Sol. Let z = x + iy. Then | z2 – 1| = | z |2 + 1 Þ | x2 – y2 – 1 + i2xy | = | x + iy |2 + 1 Þ (x2 – y2 – 1)2 + 4x2y2 = (x2 + y2 + 1)2 Þ 4x2 = 0 i.e., x = 0 Hence z lies on y-axis.

4.2 Solve following problems with the help of above text and examples. 1. For any two complex number z1, z2 | 1 - z1z 2 |2 - | z1 - z 2 |2 is equal to :

2.

(a) (1+ | z1 |2 ) (1+ | z 2 |2 )

(b) (1- | z1 |2 ) (1- | z 2 |2 )

(c) (1+ | z1 |2 ) (1- | z 2 |2 )

(d) (1- | z1 |2 ) (1+ | z 2 |2 )

If z1, z2 and z3, z4 are two pairs of conjugate complex

æ z1 ö æz numbers, then arg ç ÷ + arg çç 2 ç z4 ÷ è z3 è ø (a) 0 (c)

3p 2

(b)

ö ÷ equals ÷ ø

p 2

(d) p

3. If |b| = 1, then

b-a is equal to 1 - ab

(a) 0 (b) ½ (c) 1 (d) 2 4. Let z1 and z2 be complex numbers such that z1 ¹ z 2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (a) (b) (c) (d)

z1 + z 2 may be z1 - z 2

real and positive zero real and negative either zero or purely imaginary

2 z 1 + 3z 2 5z 2 is purely imaginary number then 2 z - 3z is 1 2 7 z1 equal to

5. If

(a)

5 7

(b)

7 9

(c)

25 49

(d) None

8

Ph y si cs

6. If z1 and z2 be two non-zero complex numbers such that

(a)

| z1 + z 2 | = | z1 - z 2 | then arg (z1) – arg (z2)= (a)

p 2

(c)

9.

p (d) 4

1

(b)

2 2

2 1

(d) None of these 3 If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then

(b) 0

p (c) 2

1

2

( a 2 + b2 ) ( c2 + d 2 ) ( e2 + f 2 ) ( g + h 2 ) = (1 + i 3 )

7.

The Argument of the complex number z =

8.

p p (b) 6 4 p (d) None of these (c) 2 The modulus of the complex number

2

is

4i(1 – i 3 )

(a)

(b) A 2 + B 2 (d) A 4 – B4 1 10. If q is real, then the modulus of is (1+ cos q ) + i sin q (a) A 2 – B 2 (c) A 4 + B 4

1 q sec 2 2 q (c) sec 2

1 q cos 2 2 q (d) cos 2

(a)

z = (1 – i 3 ) (cos q + i sin q) is 2 (1 – i) (cos q – i sin q)

(b)

ANSWER KEY 1. (b)

2. (a)

3. (c)

4. (d)

5. (d)

DE MOIVRE THEOREM These are two statements of De Moivre Theorem 1. (cosq+i sinq)n = cos nq + i sin nq, n Î I 2.

6. (a) 1 n \ (z )

ö æ p If n Î Q çç n = , q ¹ 0, p, q Î I ÷÷ then cos nq+i sinnq q ø è is one of the values of (cosq+ i sinq)n

8. (a)

9. (b)

1 n r

=

n

2.

1. Take care that (sin q + i cos q) n ¹ (sin nq + i cos nq) in fact

3.

Their images on argand diagram lie on a circle of radius r1 / n and centre origin. The points representing these roots from the vertices of a regular polygon of n sides

n

y

n

P2

P3

ö ö æ np æ np - nq ÷ - nq ÷ + i sin ç = cosç 2 2 ø ø è è

O

y¢

CUBE ROOTS OF UNITY z = (1)1 / 3 = [cos 2 kp + i sin 2 kp]1 / 3 = cos

If z = r(cos q + i sin q) and n is a positive integer, then 1 (z ) n 1 n =r

1 = [r (cos q + i sin q)] n

k = 0, 1, 2, ......, n – 1

2p 2p and + i sin 3 3

4p 4p - 1+ i 3 - 1- i 3 + i sin or 1, and . 3 3 2 2 2

[cos( 2kp + q) + i sin( 2kp + q)]

2 kp 2kp + i sin , k = 0, 1, 2 3 3

So, three cube roots of unity are 1, cos cos

1 n

x

Pn

= cos(a + b + g + .....) + i sin(a + b + g + .....)

ROOTS OF A COMPLEX NUMBER

P1

2q q

2. (cos a + i sin a)(cos b + i sin b)(cos g + i sin g )....

cos a + i sin a = cos( a - b) + i sin(a - b) 3. cos b + i sin b

10. (a)

é æ 2kp + q ö æ 2 kp + q ö ù ÷ú ÷ + i sinç êcosç è n øû ë è n ø Putting k = 0, 1, 2, ....., n – 1, we get n values which represent nth roots of complx number z PROPERTIES 1. These n roots always form a G. P. with common ratio ei 2p/

IMPORTANT RESULTS

é æp öù æp ö (sin q + i cos q) = êcosç - q ÷ + i sinç - q ÷ú 2 2 øû è ø ë è

7. (c)

2

æ -1 + i 3 ö ö æ æ ö æ ö ÷ = ç - 1 - i 3 ÷ and ç -1 - i 3 ÷ = ç -1 + i 3 ÷ As çç ÷ ÷ ç ç ÷ ç ÷ 2 2 2 2 ø è ø è è ø è ø

Laws of Motion So, we denote the non-real roots by w and w2 we write mostly

nTH ROOTS OF UNITY

- 1+ i 3 -1- i 3 and w2 = 2 2 ALGEBRAIC METHOD

Since 1 = cos 0 + i sin 0 , therefore

w=

(1)1 / n = (cos 0 + i sin 0)1 / n = (cos 2kp + i sin 2kp)1/ n

2pk 2pk + i sin ; k = 0, 1, 2, ........, (n – 1) n n

Let z = (1)1/3 Þ z 3 - 1 = 0 Þ (z - 1)(z 2 + z + 1) = 0 , which gives

= cos

-1± i 3 ,the non2 real roots , i.e., w and w2. Clearly, we can always write

i ; k = 0, 1, 2, ...., (n – 1) =e n

2kp

the roots of z as z = 1, the real root and z =

= 1, e (i 2p / n ) , e (i 4p / n) ,....., e[i 2( n -1) p / n ]

(z 3 - 1) = (z - 1)(z - w)(z - w 2 )

= 1, a, a 2 , a 3 ,...., a n -1 where a = e (i 2 p / n )

IMPORTANT CHARACTERISTICS OF w

SQUARE ROOTS OF A COMPLEX NUMBER Let z = x + iy and let the square root of z be the complex number a + ib. Then

1. 1 + w + w = 0 2

2. w = w 2 and (w2 ) = w , so w w = w3 = 1

x + iy = a + ib or

3. | w |=| w2 |= 1

5. For nay positive integer k, w3k = 1, w3k +1 = w and w3k + 2 = w2

6. For any real a, b, c; a + bw + cw2 = 0 Þ a = b = c 7. The cube roots of unity lie on a unit circle and divide the circumference into three equal parts 8. The points represented by cube roots of unityyform the vertices of an equilateral triangle. i w 9. | wz |=| w || z |=| z | arg( wz ) =

2p + arg( z ) 3

x¢

1 x

–1 w2

–i y¢

SOME USEFUL IDENTITIES (i)

x 2 + y 2 = ( x + iy)( x - iy)

(ii)

x 3 + y 3 = ( x + y)(x + wy)(x + w 2 y)

( x + iy) = (a + ib) 2 = (a 2 - b 2 ) + (2ab)i

Equating real and imaginary part, we get

4p or - 2p 4. arg( w) = 2 p and arg(w 2 ) = 3 3 3

10.

9

= ( x + y)(xw + yw2 )(xw 2 + yw)

(iii) x 3 - y 3 = (x - y)(x – wy)( x – w2 y) (iv) x2 + xy + y2 = (x – yw) (x – yw2), in particular, x2 + x + 1 = (x – w) (x – w2) (v) x2 – xy + y2 = (x + yw) (x + yw2), in particular, x2 – x + 1 = (x + w) (x + w2) (vi) x2 + y2 + z2 – xy – xz – yz = (x + yw + zw2) (x + yw2 + zw) (vii) x 3 + y 3 + z 3 – 3xyz = (x + y + z )(x 2 + y 2 + z 2 – xy – yz – zx)

= ( x + y + z)(x + yw + zw2 )(x + yw2 + zw)

x = a 2 - b2 and y = 2ab

…(1) …(2)

Now, a 2 + b 2 = (a 2 - b 2 ) 2 + 4a 2 b 2 = x 2 + y 2 …(3) Solving the equations (1) and (3), we get æ x 2 + y2 + x ö ÷ ç a=± ç ÷; b=± 2 ÷ ç ø è

æ x 2 + y2 - x ö ÷ ç ÷ ç 2 ÷ ç ø è

From (2), we can determine the sign of ab. If ab > 0, then a and b will have same sign. Thus é ê x + iy = ± ê ê ë

æ x 2 + y2 + x ö ÷ ç ÷ +i ç 2 ÷ ç ø è

æ x 2 + y2 - x ö ù ÷ú ç ÷ú ç 2 ÷ú ç øû è

If ab < 0, then é æ ö æ x 2 + y2 - x ö ù 2 2 ê ç x +y +x÷ ÷ú ç x + iy = ± ê ç ÷ú ÷ -i ç 2 2 ÷ú ÷ ç ç ê è øû ø è ë Thus, square roots of z = a + ib are :

é | z | +a | z | -a ù ±ê +i ú 2 2 ûú for b > 0 ëê and

é | z | +a | z | -a ù ±ê -i ú 2 2 ûú for b < 0 ëê

For example : æ 1+ i ö (i) Square root of i is ± çç ÷÷ , as x = 0, y = 1 > 0 and è 2 ø |i|=1

10

Ph y si cs

æ1- i ö ÷÷ , as x = 0, y = -1 < 0 and | i | = (ii) Square root of -i is ± çç è 2ø

1 æ 5-3 5 - ( - 3) (iii) Squar e root of -3 + 4i is ± ç +i ç 2 2 è

ö ÷, ÷ ø

or ± (1 + 2i) , As x = –3 and y = 4 > 0.

æ 3 -iö ÷ ; As y = - 3 < 0 ±ç ç 2 ÷ è ø (-1 + i 3 )15

13.

(1 - i ) 20

+

(1 - i ) 20 15

[

=

(-2i)

10

+

(1 + i) 20

æ -1- i 3 ö ÷ 215 ç ç ÷ 2 è ø +

]

215 (w)15

(-1 - i 3 )15

15

æ -1+ i 3 ö ÷ 215 ç ÷ ç 2 ø è = 2 10 (1 - i)

c + a w + bw 2 (a) –1

é 1 w3 =ê = = w2 êë w w

10

b + cw + aw

2

= w2 + w = -1

ù ú úû

Answer (a)

å a r cos r x , then

r=0

25 (w3 )5

(2i )

2 5

(i )

a + bw + c w 2 b + cw + aw2 (b) 1

(b) a0 = 0

r=0

(c) a6 = 1 (d) a1 = a3 = a5 Sol. [To find the value of cos nx and sin nx in ascending powers of cos x or sin x we expand (cosx+isinx)n using De Moivre theorem and Binomial theorem then equate real and imaginary parts from two expansion to get the required identity as done for above example]. We have (cosx + i sinx)6 = cos6x + isin6x ...(i) (Using De Moivre theorem) Also (cos x + isinx)6

[(1 + i) ]

=

etc.

w(aw 2 + b + cw)

å ar = 1

+ 6 C 3 cos 3 x i 3 sin 3 x + 6 C 4 cos 2 x i 4 sin 4 x

2 10

215 (w2 )15

+

c + aw + bw

2

+

= 6 C cos 6 x + 6 C cos 5 x.i sin x + 6 C cos 4 x i 2 sin 2 x 0 1 2 + 6 C 5 cos x i 5 sin 5 x + 6 C 6 i 6 sin 6 x

+

25 (w3 )10 2 5

(i )

Answer (c) = -2 5 - 2 5 = -64 [You can also convert the numbers into polar form and apply De Moivre theorem but above approach is better] 14. The value of (2 – w) (2 – w2) (2 – w10) (2 – w11) is (a) 49 (b) 16 (c) 16w (d) 49w2 10 9 11 9 2 2 Sol. w = w .w = w, w = w .w = w \ (2 – w) (2 – w2) (2 – w10) (2 – w11) = (2 – w)2 (2 –w2)2 Now x3 – 1 = (x – 1) (x – w) (x – w2), dividing by x – 1 \ x2 + x + 1 = (x – w) (x – w2) or (x2 + x + 1)2 = (x – w)2 (x – w2)2. Put x = 2, (2 – w)2 (2 – w2)2 = (7)2 = 49 Answer (a) 15. If w is a complex cube root of unity, then the value of a + bw + cw 2

w 2 (aw + bw 2 + c)

(a)

is equal to

(1 + i) 20

+

b + cw + aw2

6

(-1 - i 3 )15

(-1 + i 3 )15

Let z =

a + bw + c w 2

æ a ö b a w2 çç + + c ÷÷ wæç + b + cw ö÷ 2 w w èw ø+ è ø = b + cw + aw2 c + aw + bw 2

16. If cos6x =

3 : 1 can be expressed in terms of w and w2)

3 or

c + a w + bw 2

+

6

(a) 132 (b) 64 (c) – 64 (d) none Sol. (Trick :Any complex number z with |Re(z) | : | Im(z) | = 1 :

a + bw + cw 2

=

æ 2 +1 2 - 1 ö÷ or -i 1- 3 i is ± ç ç 2 2 ÷ø è

(iv) Square root of

Sol.

(Using Binomial theorem) = 6 C 0 cos 6 x - 6 C 2 cos 4 x sin 2 x + 6 C 4 cos 2 x sin 4 x -6 C6 sin 6 x 6 5 6 3 3 6 5 + i[ C1 cos x sin x - C3 cos x sin x + C5 cos x sin x] From (i) & (ii), equating the real parts we get

Cos 6x = 6 C 0 cos 6 x - 6 C 2 cos 4 x sin 2 x + 6 C 4 cos 2 x sin 4 x - 6 C 6 sin 6 x 6 4 2 2 2 2 = cos x - 15 cos x (1 - cos x ) + 15 cos x(1 - cos x)

- (1 - cos 2 x )3 6

= 32 cos6x – 48 cos4x + 18 cos2x – 1=

å a r cos 6 x

r =0

We get, a0 = – 1, a1 = 0, a2 = 18, a3 = 0, a4 = – 48, a5 = 0, a6 = 32 6

is Clearly (c) a + b + c (d) 0

...(ii)

å a r = a 0 + a1 + a 2 + a 3 + a 4 + a 5 + a 6 = 1 .

r =0

Also a1 = a3 = a5 = 0.

Answer (a, d)

11

Laws of Motion 5

å a r sin r x , then

17. If sin 5x = 5

(a)

å

r =0

= -

1 [2 cos 6x - 6 ´ 2 cos 4x + 15 ´ 2 cos 2x - 20] 64

= -

1 3 15 5 a r cos rx cos 6 x + cos 4 x - cos 2 x + = 32 16 32 8 r =0

r =0

6

ar =1

(b) a1 = a3 = a5

(c) a0 = a2 = a4 (d) a1+a3+a 5=a0+a2+a 4 Sol. Solve as above, expand (cos x + i sinx)5 using two theorems and equate imaginary parts. Answer (a, c)

å

6

18.

If sin 6x

=

å a r cos rx . Then

and a6 = -

r =0

(a) a0 = 0

(b) a1 = a3 = a5

1 2 (d) 2a0 + a1 + 3a2 + a3 + a4 + a5 + a6 = 0 Sol. To express cosnx or sinnx as a series of multiple angles of cos and sin, we use Euler's representation cosx =

5 15 3 , a = 0, a2 = , a = 0, a4 = ,a =0 8 1 32 3 16 5

We get a0 =

1 . 32

Only (d) is satisfied with these values. Answer (d) 19. If a, b, g are the cube roots of a negative number p then

(c) a2 + a6 =

e ix + e -ix e ix - e -ix and expand cosnx = and sinx = 2 2i æ e ix + e -ix ç ç 2 è example.

[Ce

1

6

6 6

2 i

0

6ix

ö ÷ ÷ ø

- 6 C1e 5ix .e -ix + 6 C 2 e 4ix e - 2ix

- 6 C5 e ix e -5ix + 6 C6 e -6ix

=

- 26

[ C (e 6

(a)

- 1- 3i 2

(b) (x + y + z)i

0

6ix

+e

æx y ö (d) ç + i ÷p è2 2 ø

(c) ip

Sol. Let q = –p, q > 0 Cube root of p = p1/3 = (–q)1/3 = –q1/3. (1)1/3 Cube roots of p are – q1/3, –q1/3w, –q1/3w2 Let a = –q1/3, b = –q1/3w, g = –q1/3w2

6

- 6 C3e 3ix e -3ix + 6 C 4 e 2ix e - 4ix

1

-6ix

) - C1 (e

+ C 2 (e 6

- 4ix

6

4ix

+e

2ix

- 2ix

) - C3

+e

6

]

)

xa + yb + zg x (-q1 / 3 ) + y(-q1 / 3 w ) + z(-q1 / 3 w 2 ) \ xb + yg + za = x (-q1 / 3 w ) + y(-q1 / 3 w 2 ) + z (-q1 / 3 ) x + yw + zw 2

=

xw + yw + z 2

=

w 2 ( xw + yw 2 + z ) xw + yw + z 2

= w2 =

]

Answer (a)

4.3 Solve following problems with the help of above text and examples. ( 3 + i)

2. Arg

17

(a)

(1 – i )50

2p 3

i ( 3 + i )6 4(1 – i 3 ) 2

, then amp (z) is equal to p 3

(c) - 2p 3

(d) None (c)

is equal to

p 6

(b)

p 6

3p 10

(d)

5p 10

(a) – (b) –

- 1 - 3i 2

æ ö or ç -1 + 3i ÷ è 2 ø

[Qn C r =n C n - r ]

1. If z =

xa + yb + zg xb + yg + za

is

n

ö ÷ by binomial theorem etc. as for the above ÷ ø

æ e ix - e -ix sin6x = çç 2i è

=

for any three real numbers x, y, z, the value of

12

Ph y si cs

3. For z = cos q + i sin q, then the value of

1 - z 2n

If w ¹ 1 is a cube root of unity, then the roots of the equation (x+aw)3 + a3 = 0 are :

8. is

1 + z 2n (a) –i tan n q (b) i tan n q (c) tan n q (d) i 4. If x = a + b, y = aw + bw2 and z = aw2 + bw, then x3+y3+z3 = (a) 3 (a3 + b3) (b) (a + b)3 3 3 2 2 (c) a + b – a b – ab (d) None of these 2 5. If 1, w, w are the cube roots of unity then (1 – w + w2)5 + ( 1 + w – w2)5 = (a) 32 (b) 0 (c) 32w (d) 16w2 2 6. If 1, w, w are the cube roots of unity then (3 + 3w + 5w2)6 – (2 + 6w + 2w2)3 = (a) 0 (b) 64 (c) 36 (d) – 36

(a) a, aw2, – 2aw (c) –2a, aw2, aw

Let a = z1 + z 2 , b = wz1 + w2 z 2 , g = w2 z1 + wz 2 where w is a complex cube root of unity, then :

9.

(a) a 2 + b 2 + g 2 = 8 z1z 2 (b) ab + bg + ga = 3z1z 2 (c) a 3 + b 3 + g 3 = 3( z13 + z 2 3 ) (d) abg = 2(z13 + z 2 3 ) 10. The polynomial x 3m + x 3n +1 + x 3k + 2 , is exactly divisible by x 2 + x + 1 if (a) m n, k are rational (b) m, n, k are integers (c) m, n, k are positive integers (d) none of these.

7. x Î R , then square roots of x + i x 4 + x 2 + 1 are (a) ± (b) ± (c) ± (d) ±

1 æ 2 ö 2 ç x + x + 1 + i x - x +1 ÷ ø 2è

(b) a, –2aw2, aw (d) a, – 2aw2, –aw

334

æ ö 11. If i = - 1, then 4 + 5 ç - 1 + i 3 ÷ ç ÷

1 æ 2 ö 2 ç x + x +1 - i x - x + 1 ÷ ø è 2

è 2

2 ø

365

æ 1 i 3ö + 3 çç - + ÷ 2 ÷ø è 2

is

equal to

1 æ 2 ö 2 ç x - x + 1 + i x + x +1 ÷ ø 2è 1 æ 2 ö 2 ç x - x +1 - i x + x +1÷ ø 2è

(a) 1- i 3

(b) - 1+ i 3

(c) i 3

(d) - i 3

12. If w (¹ 1) be a cube root of unity and (1 + w) 7 = A + Bw, then A and B are respectively the numbers (a) 0, 1 (b) 1, 1 (c) 1, 0 (d) –1, 1 ANSWER KEY

1. (c)

2. (b)

3. (a)

4. (a)

5. (a)

6. (a)

7. (a)

8. (a)

9. (c)

10. (b) 11. (c) 12. (b)

GEOMETRY OF A COMPLEX NUMBER Y

SECTION FORMULA

As in vectors, we represent a point by the position vector of the

OP= z

P (z1)

P(z)

point OP = r with r esp ect t o orgin O. Similarly the point P can be repr esen ted by a complex n umber z, su ch t h at l engt h

R (z) O

X

PR m = QR n

Q (z2)

and ÐXOP = arg(z ) . The point P is called the IMAGE of the complex number z and z is said to be AFFIX or complex coordinate of the point P. DISTANCE BETWEEN TWO POINTS If two points P and Q have affices z1and z2 respectively then

If a point R (z) divides the join of two points P (z 1) and Q (z2) in ratio m:n, then z=

mz 2 + nz1 mz 2 - nz1 (internally) and z = (externally) m+n m-n

PQ = z 2 – z1

Ø

Mid Point of PQ is given by

PQ = z 2 – z1 = Affix of Q – Affix of P..

z1 + z 2 2

Laws of Motion ANGLE BETWEEN TWO LINES (CONCEPT OF ROTATION)

z3 – z4 Then z – z is purely imaginary.. 1 2

Þ

R (z3 ) Q (z2 )

q

z3 – z 4 z – z4 =– 3 z1 – z 2 z1 – z 2

or alternatively

P (z1 ) q1

(i)

13

z3 – z 4 = ik, for some k Î R - {0} z1 – z 2

Þ z 3 – z 4 = ik (z1 – z 2 ) (iii) Multiplying a complex number z by i is equivalent to rotating the image of z in Argand plane by 90º about origin,

q2

Consider three points P(z1), Q(z2) and R(z3) Then angle between PQ an PR (counter clockwise)

p æ iz ö anticlockwise, as |z| = |iz| and arg ç ÷ = arg (i) = 2 èzø

q = q 2 - q1 = arg( PR ) – arg( PQ ) = arg( z 3 – z 1 ) – arg( z 2 – z 1 )

Q (iz)

z –z \ q = arg 3 1 z 2 – z1

P (z)

If P(z1), Q(z2) and R(z3) be collinear Points, then q = 0 or p,

p/ 2

O

z1

i.e.

z 3 – z1 z – z = 3 1Þz 2 z 2 – z1 z – z 2 1 z3

z1 1 z2 1 = 0 z3 1

(A complex number z is purely real if z = z ) If PR is perpendcular to PQ.Then

(iv) Multiplying a complex number z by w is equivalent to rotating the image of z in Argand plane by 120º (or 240°) about origin anticlockwise, for |z| = |wz| (|w| = 1) and

2p æ 4 p ö æ wz ö arg ç ç or ÷ ÷ = arg (w) = 3 è 3 ø è z ø

æz –z ö z –z p arg çç 3 1 ÷÷ = ± Þ 3 1 is puerly imaginary 2 z 2 – z1 è z 2 – z1 ø

That is,

z3 – z1 z –z =- 3 1 z 2 – z1 z2 – z1

P (z)

Q (wz) 2p 3

(A Complex number z is purely imaginary if z = - z ) (ii)

O

R (z 3 )

P (z 1 )

ANGLES OF A TRIANGLE

q

C (z3 ) Q ( z2 )

S (z4 )

B (z 2 ) A (z1 )

Consider Four Points P, Q, R and S with affices respectively z1, z 2 , z 3 and z4. As in (i) the angle between SR and QP.. æ z – z4 q = arg çç 3 è z1 – z 2

If z1 z2 and z3 be the affices of vertices A, B and C of a tringle ABC described in counterclockwire sense. Then

ö ÷ ÷ ø

If SR and QP be perpendicular, q = ±

p 2

z 3 – z1 CA = (cos A + i sin A) or z 3 – z1 = CA e iA z 2 – z1 BA z 2 – z1 BA Similarly relations with other vertical angles can be given.

14

Ph y si cs IMPORTANT RESULTS ABOUT TRIANGLES

If z1, z2 , z3 are the vertices of a triangle then 1. centroid z is givn by z =

z1 + z 2 + z 3 3

2. Incentre I (z) of the DABC is given by z =

EQUATION OF STRAIGHT LINE THROUGH TWO POINTS Z1 AND Z2 Let variable point z be a point on this line then z

az1 + bz 2 + cz 3 a +b+c

3. Circumcentre O(z) of theDABC is given by

z1 z2

Þ z ( z1 - z 2 )i + z ( z 2 - z1 )i + i(z1z 2 - z1z 2 ) = 0

(i) | z1 |2 | z 2 |2 | z 3 |2

z=

z12 z22 z 32

z1 1 z2 1 z3 1 z1 1 z2 1 z3 1

2

=

z1 (sin 2A ) + z 2 (sin 2 B) + z 3 (sin 2C) sin 2A + sin 2 B + sin 2C

z2 z3 2

z=

| z1 |2

z1 1

z 2 1 + | z 2 |2 z 3 1 | z 3 |2

z2 1 z3 1

z1 1

z12 z22 z 32

Let (z 2 - z1 )i = a , a constant complex number then

-(z 2 - z1 )i = a Also i (z1z 2 - z1z 2 ) is purely real constant

4. Orthocentre H(z) of DABC is given by z12

z 1 z1 1 = 0 or z( z1 - z 2 ) + z (z 2 - z1 ) + z1z 2 - z1z 2 = 0 z2 1

number, say b then the above equation is az + az + b = 0 It is called the general equation of a straight line (ii) The complex equation za + z a + b = 0 represents a straight line in complex plane where 'a' is a complex number and 'b' is a real number. The complex slope of the line is given a a

by - . (iii) The equation of the perpendicular bisector of the line segment joining the points A(z1) and B(z2) is

OR

z ( z1 - z 2 ) + z (z1 - z 2 ) =| z1 |2 - | z 2 |2

z1 1 z2 1 z3 1

IMPORTANT RESULTS 1. The complex slope of line joing points A (z1) and B(z2)

(tan A)z1 + (tan B)z 2 + (tan C)z3 tan A + tan B + tan C (a sec A)z1 + (b sec B)z 2 + (c sec C)z3 = a sec A + b sec B + csec C 5. The centroid G lies on the segment joining the orthocentre H and the circumcentre O of the triangle and divides z=

HG 2 = internally in ratio 2 : 1, i.e OG 1

z1 - z 2 is define as m = z - z 1 2

2. Two lines with complex slopes m1 and m 2 are parallel if m1 = m 2 and perpendicular if m1 + m 2 = 0 3. The length of perpendicular from a point A(a) to the line az + az + b = 0 is given by p =

| aa + a a + b | 2|a|

H G

O 6. Area of the DABC is given by the modulus of

1 4

EQUATION OF A CIRCLE :

z1 z1 1 z 2 z2 1 z3 z3 1

P (z) r C (z 0 )

7. Triangle ABC is equilateral if and only if 1 1 1 + + =0 z 2 - z 3 z 3 - z1 z1 - z 2 Û z12 + z 22 + z 32 = z 2 z 3 + z 3 z1 + z1z 2

1 z2 Û 1 z3 1 z1

z3

(1) Consider a circle with centre C having affix z0 and radius r. For any point P(z) on this circle CP = r

z1 = 0

i.e. z – z 0 = r

z2

where, r is a positive real number

.....(1)

Laws of Motion

Ø

z – z 0 < r represents the region inside the circle given by (1)

Ø

z – z 0 > r represents the region outside the circle given by (1). (2) General equation of a circle Consider the equation of a circle | z - z 0 |= r Þ ( z - z 0 )( z - z 0 ) = r 2 Þ zz - zz 0 - zz 0 + | z 0 |2 -r 2 = 0

Let -z 0 = a , a constant complex number and | z 0 | 2 -r 2 =| a |2 -r 2 = b, a constant real number then the

above equation becomes zz + az + az + b = 0 , It is called the general equation of circle . Hence the complex equation z z + a z + az + b = 0 reprectents a circle in complex plane where 'a' is a complex number and 'b' is a real number. The centre of the circle is a point with affix '–a' and the radius is given by

a

2

–b.

For the existence of the circle a 2 – b > 0. (3) (z – z1 )(z – z 2 ) + (z – z 2 ) (z – z 1 ) = 0 represents a circle in the complex plane which is described on a line as diameter having extremities z1 and z2. 20. Let z1 = 10 + 6i and z2 = 4 + 6i. If z is any complex number such that the argument of

z - z1 p is . Then z must lie on z - z2 4

18

(c) Centre ( 2 , 1); radius 18 (d) Centre (7, 9); radius 6 Sol. Let z = x + iy Then =

6 y - 36 2

2

x + y - 14 x - 12 y + 76

Þ + y2 – 14x – 18y + 112 = 0 which is a circle with centre (7, 9)

x 2 + y 2 - 8x - 12 y + 52

+

..... (i)

7 2 + 9 2 - 112 = 49 + 81 - 112 = 18 Answer (b) Note : The equation (i) may be converted to complex form as following : x2 –14 + y2 – 18y + 112 = 0 Þ (x2 – 14x + 49) + (y2 – 18y + 81) + 112 – 49 – 81 = 0 Þ (x – 7)2 + (y – 9)2 = 18 Þ |(x – 7) + i (y – 9)|2 = 18 Þ |(x + iy) – (7 + 9i)|2 = 18 and radius =

Þ |z – (7 + 9i)|2 = 18 Þ |z – (7 + 9i)| = 18 , which is equivalent to |z – z0| = r Hence centre of the circle is z0 = 7 + 9i = (7, 9) radius of the circle is r =

18 21. Consider the complex number z satisfying |z – 5i| £ 3, then (a) Value of z having the least modulus is z =2i (b) Value of z having the greatest modulus is z = 8i (c) Value of z having the least positive argument is 4 (3 + 4i) 5 (d) Value of z having the greatest positive argument is

z=

z=

4 (-3 + 4i) 5

z - z1 ( x + iy ) - (10 + 6i ) ( x - 10 ) + ( y - 6) i = = z - z2 ( x + iy) - (4 + 6i ) ( x - 4) + ( y - 6) i

x 2 + y 2 - 14x - 12 y + 76

p =1 4

x2

y

(x - 10) + ( y - 6) i (x - 4) - ( y - 6)i x (x - 4) + ( y - 6) i (x - 4) - ( y - 6)i

=

= tan

Sol. The inequation | z – 5i | £ 3 represents all the complex numbers lying inside or on the circle | z – 5i | = 3 ...(1) Clearly the circle (1) has centre at (0, 5) and radius = 3 Let us plot this circle on the xy – plane. The given inequation represents the points inside or on the circle shown in figure.

a circle, which has (a) Centre (0, 0); radius 6 (b) Centre (7, 9); radius

Þ

15

E

C (5i)

(6 y - 36)i x 2 + y 2 - 8x - 12y + 52 (on simplifying)

æ z - z1 ö p Now given that arg çç z - z ÷÷ = 4 ; 2ø è é ù p 6y - 36 Hence tan -1 ê 2 2 ú= êë x + y -14x -12y + 76 úû 4

é -1 y ù êQ arg ( x + iy) = tan x ú ë û

B D b

A f a

x

O

Note that the modulus of a complex number is represented by the distance of the image of the complex number from origin. Clearly the point at the least distance from origin is D and the point at the greatest distance from origin is E. Hence the affices of D and E give us the complex numbers of the least and the greatest moduli respectively.

16

Ph y si cs

Now OC = 5, DC = 3, Hence OD = 2 and OE = 8 \ Point D, i.e. complex number with least modulus is 2i Point E, i.e. complex number with maximum modulus is 8i Furthermore the argument of a complex number in Argand diagram is given by the angle that the line joining the origin to its image forms with positive x–axis. Clearly the two tangents drawn on the circle from origin, OA and OB represent the least and the greatest values of this angle respectively for any point on or inside the circle. Hence, the affices of points A and B give us the complex numbers with least and greatest arguments respectively. Now Ð XOA = a = and ÐXOB = b = Clearly cos f =

(c)

(d) z12 + z 22 + z 32 - z1z 2 - z 2 z 3 - z 3 z1 = 0 Sol. Let a = z1 - z 2, b = z 2 - z 3 and g = z 3 - z1 Then a + b + g = 0 Clearly a + b + g = 0 Let the triangle be equilateral, then | z1 – z2 | = |z2 – z3 | = | z3 – z1 | = l say That is | a | = | b | = | g | = l

p - f, f = ÐAOC 2

Now a a = | a | 2 = l2 Þ a =

p +f 2 OA = OC

OC 2 - AC2 = OC

[Where AC = radius = 3 ] and sin f =

1 1 = z1 - z 2 z 2 - z 3

Similarly, b =

5 2 - 32 4 = 5 5

Affix of point A is r (cos a+ i sin a)

Þ

ö æp 3 r = OA = 4, cos a = cos ç - f ÷ = sin f = ; 5 ø è2

....(2)

l2 a

l2 l2 &g = b g

Hence, from (2)

3 5

....(1)

l2 l2 l2 1 1 1 + + =0 Þ + + =0 a b g a b g

1 1 1 + + =0 z1 - z 2 z 2 - z 3 z 3 - z1

1 1 1 Conversely let z - z + z - z + z - z = 0 1 2 2 3 3 1

æp ö 4 sin a = sin ç - f ÷ = cos f = . \ A is 4 (3 + 4i) 5 è2 ø 5 Also affix of point B is r(cos b + i sinb)

Þ

ö æp 3 r = OB = OA = 4, cos b = cos ç + f ÷ = – sin f = – 2 5 ø è

1 1 1 1 b+g a + + =0Þ = = a b g a bg bg

from (1)

\a 2 = by Þ | a |2 =| bg |Þ| a |3 =| a || b || g | 3 Similarly | b | =| a || b || g | and | g |3 =| a || b || g |

ö æp 4 sin b = sin ç + f ÷ = cos f = 2 5 ø è

Hence | a | = | b | = | g | Þ | z1 - z 2 | = | z 2 - z 3 | = | z 3 - z1 | . That is the triangle is equilateral.

4 (-3 + 4i) 5 Hence, the complex number with least argument is

Also note that

\ B is

4 (3 + 4i) and the complex number with greatest argument 5

4 (-3 + 4i) . 5 All options (a), (b), (c) and (d) are correct Note : Student may feel that the solution of this example is lengthy on the contrary the solution is quite simple and direct from the figure only. 22. Suppose that z1, z2, z3 represent the vertices of a triangle taken in order. The triangle is equilateral if and only if

Þ z12 + z 22 + z 32 - z1z 2 - z 2 z 3 - z 3 z1 = 0 Answer (a, d) 23. The centre of a regular hexagon has affix i. The affix of one vertex is 2 + i. The affix z of adjacent vertices are

is

(a)

1 1 1 + + =0 z1 - z 2 z 2 - z 3 z 3 - z1

(b)

1 1 1 =0 z1 - z 2 z 2 - z 3 z 3 - z1

1 1 1 + =0 z1 - z 2 z 2 - z 3 z 3 - z1

Sol.

(a) 1 + i (1 ± 3 )

(b) i + 1 ± 3

(c) 2 + i (1 ± 3 )

(d) 1± 2i

Laws of Motion p Let A be the image of complex number 2 + i, ÐAOB = 3

z-i Let B (or F) be z, then ÐAOB = arg (2 + i) - i (refer to angle between two lines in the text) =±

\

p 3

25. Let z1 and z2 be the roots of complex equation z2 + pz + q = 0. The points represented by z1, z2 and the origin form an equilateral triangle if (a) p2 > 3q (b) P2 < 3q (c) p 2 = 3q (d) p = 3q 2 Sol. Given z1 and z2 are roots of z + pz + q = 0 Hence, z1 + z2 = – p – (i), z1z2 = q –(ii) We know that if z1, z2, z3 are vertices of an equilateral triangle then z12 + z 22 + z 32 - z1 z 2 - z 2 z 3 - z 3 z1 = 0 (see example 16)

z -i BO æ p pö = ç cos ± i sin ÷ ( 2 + i) - i AO è 3 3ø

= cos

17

p p ± i sin 3 3

Here z3 = 0, then z 12 + z 22 - z 1z 2 = 0

[Q AO = BO]

æ1 3ö \ z - i = 2 çç ± i ÷÷ Þ z = i + 1 ± i 3 = 1 + i 1 ± 3 2 ø è2

(

)

Answer (a) 24. Let a , b Î R , such that 0 z1 - z 2 , 'a' is positive real number represents an ellipse in complex plane, z1 and z2 are affices of two foci of ellipse. If 2a = | z1 - z 2 |, then | z - z1 | + | z - z 2 |= 2a represents the line segment joining z1 and z2 If 2a 0, x 2 + y 2 - 1 > 0,

5 5 (3 + 4i) or (3 + 4i ) (b) 3 7

2y 2

x + y2 - 1

=1

Þ x 2 + y 2 - 2 y - 1 = 0, y > 0, x 2 + y 2 - 1 > 0

(d) None of these.

Therefore the locus represented by the equation æ z -1 ö p ÷= arg ç è z +1 ø 4 is the arc ABC of the circle x2 + y2 – 2y – 1 = 0. Solving with x = 0, we get

Sol.

y=

2± 8 = 1 ± 2 , y > 0,\ y = 1 + 2 2

Answer (b)

31. In argand plane the locus of z ¹ 1 such that | z – a | = 2 is the circle with centre at a = (3, 4) and radius = 2. Clearly ÐXOC = q = tan –1

4 3

Since arg (z) = tan –1 4 , z must lie along OC and on the 3 circle. It must therefore be either point A or B. Now, OA = OC – AC = 5 – 2 = 3 OB = OC + CB = 5 + 2 = 7

7 B is 7 (cos q +i sin q ) = (3 + 4 i) 5

(a) (b) (c) (d) Sol.

the straight line joining the points z = 3/2, z = –2/3 the straight line joining the points z = –3/2, z = 2/3 a segment of a circle passing through z = 3/2, z = –2/3 a segment of a circle passing through z = –3/2, z = 2/3

2 z 2 - 5z + 3 3z - z - 2 2

=

(2z - 3)( z - 1) 2z - 3 2 z - 3 / 2 as z ¹ 1. = . = (3z + 2)(z - 1) 3z + 2 3 z + 2 / 3 æ z - 3/ 2 ö

2p

\ The given condition reduces to arg ç z + 2 / 3 ÷ = 3 è ø

3 \ A is 3 (cos q +i sin q ) = (3 + 4i) 5

This implies that the line joining the points z = Answer (a)

z -1 p = 30. If z = x + iy such that | z + 1| = | z – 1| and amp z +1 4 then

(a) x = 2 + 1, y = 0

(b) x = 0, y =

(c) x = 0, y =

(d) x =

2 -1

é 2z 2 - 5z + 3 ù 2p ú= arg ê 2 3 is êë 3z - z - 2 úû

z= -

2 - 1, y = 0

2 2p subtend a constant angle= at the point z. Thus 3 3

z describes the segment of a circle through z =

2 +1

z= -

3 . and 2

3 and 2

2 2p at which the chord subtends an angle= . 3 3 Answer (c)

Laws of Motion

21

4.5 Solve following problems with the help of above text and examples. 1. If z = x + iy and w =

1 - iz , then |w| = 1 implies that, in the z-i

complex plane. (a) z lies on the imaginary axis (b) z lies on the real axis (c) z lies on the unit circle (d) None of these. 2. The Points representing the complex numbers z for which | z + 4 |2 – | z – 4 |2 =8 lie on (a) a straight line parallel to x – axis (b) a straight line parallel to y – axis (c) a circle with centre as origin (d) a circle with centre other than the origin 3. If z1 , z 2 , z 3 are in H.P. they lie on a: (a) circle (b) sphere (c) straight line (d) None of these 4. The locus of the point Z in the Argand plane for which Z +1

2

+ Z –1

2

6. If the real part of

representing z in the complex plane is a/an (a) circle (b) parabola (c) hyperbola (d) ellipse The complex number z = x + iy which satisfy the

7.

equation

(b) pair of straight lines (d) parabola

2z + 1 5. If the imaginary part of is – 2, then the locus of z in iz + 1

the complex plane is (a) a circle (c) a parabola

(b) a straight line (d) None of these

z - 5i = 1 lie on z + 5i

(a) the x -axis (c) a circle through the origin

(b) the line y = 5 (d) None of these

z-i = 1 , then locus of z is z +1

8.

If

9.

(a) x-axis (b) y-axis (c) x = 1 (d) x + y = 0 If z = x + iy and ‘a’ is a real n umber such that | z - ai | = | z + ai | , then locus of z is

= 4 is a

(a) straight line (c) circle

z+2 is 4, then the locus of the point z –i

(a) x-axis

(b) y - axis

(c) x = y

(d) x 2 + y 2 = 1

10. If | z + z | + | z - z |= 8, then z lies on (a) a circle (b) a straight line (c) a square (d) None of these 11. If z is a complex number satisfying |z – i Re(z)| = |z – Im(z)| then z lies on (a) y = x (b) y = - x (c) y = x + 1

(d) y = - x + 1

ANSWER KEY 1. (b)

2. (b)

3. (a)

4. (c)

5. (b)

6. (a)

7. (a)

8. (d)

9. (a)

10. (c) 11. (a, b)

SOME IMPORTANT INEQUALITIES 1.

z1 | – | z 2 is the least value of | z1 + z2 | and z1 + z 2 is the greatest value of z1 + z 2

Thus z1 - z 2 £ z1 + z 2 £ z 2 + z 2 2. - | z | £ Re( z ) £ | z | 3. - | z | £ Im(z) £ | z | 4. | z | < | Re(z) | + | Im(z) | < 5. If z +

2 |z|

2 2 1 = a , a is positive real number then - a + a + 4 £ | z | £ a + a + 4 z 2 2

22

Ph y si cs

1.

If the complex number z satisfies the equation iz3 + z2 – z + i = 0, then (a) z lies on a unit circle (b) Re(z) = Im (z) (c) | z | = 1 (d) None of these Sol. iz3 + z2 – z + i = 0 Þ i2z3 + iz2 – iz + i2 = 0 (Multiply with i) Þ – z3 + iz2 – iz + i2 = 0 (i2 = –1) Þ –z(z2 + i) + i (z2 + i) = 0 Þ (z2 + i) (z – i) = 0 Either z = i Þ |z| = 1 or z2 = –i Þ |z2| = 1 Þ |z| = 1 Hence, |z| = 1, implies that z lies on unit circle. Answer (a, c) 2. All non zero complex numbers z satisfying the equation z = iz 2 are (a) i

(b) –i

3 1 3 1 - i - i (d) 2 2 2 2 Sol. Let z = x + iy, x, y Î R [Note that the problems with lower powers of z can easily be solved using z = x + iy] (c)

Then z = iz 2 Þ x – iy = i (x + iy)2 Þ x – iy = i (x2 – y2 + 2ixy) = i (x2 – y2) – 2xy Equating real and imaginary parts x = –2xy and –y = x2 –y2

Sol.

Using the result |z1+z2 | £ |z1| + |z2|, we get |z| + |z – i| = |z| + |i – z| [since |z| = | – z|] £ | z + i – z | = |i| = 1 \ minimum value of |z| + |z–i| is 1 ALTERNATE : We may obtain the above answer using geometrical representation. Consider a triangle which has vertices O (origin) P (z) and Q (z – i) then OP = |z|. OQ = | z – i| and PQ | z + i–z| = |i| = 1 Now in a triangle sum of two sides ³ third side That is OP + OQ ³ PQ [The equality holds, when O, P and Q are collinear] Thus | z | + | z – i| ³ 1 Answer (b) 4.

(a)

1 1 3 3 = x2 - Þ x2 = Þ x = ± 2 4 4 2 Discording the solution x = 0, y = 0 (for it is z = 0) We get the solution z = 0 + 1i = i,

3.

3 1 - 3 1 - i and z = - i Answer (a, c & d) z= 2 2 2 2 The minimum value of |z| + | z – i| is (a) 0 (b) 1 (c) 2 (d) none

2

(b)

3 7 4

(c) 2

(d) None

Þ z = i, 1 - i, – 1 – i Let A = (0,1), B = (1,–1), C=(–1, –1),

1 2 If x = 0, from second equation, y (y – 1) = 0 Þ y = 0 or 1 1 from second equation 2

3 7 2

3

Sol. z 3 + iz 2 + 2i = (z - i )(z 2 + 2iz - 2) = 0

x = – 2xy Þ x (1 + 2y) = 0 Þ x = 0 or y = -

If y = -

If the roots of z + iz + 2 i = 0 represent the vertices of a DABC in the Argand plane, then the area of the triangle is

0 1 1 1 1 1 - 1 1 = | - 2 - 2 |= 2 \ D ABC = 2 2 - 1 -1 1

5.

Answer (c)

The complex number z satisfying the equations | z | -4 =| z - i | - | z + 5i |= 0 , is

(b) 2 3 - 2i (c) - 2 3 - 2i 3 -i Sol. We have two equations | z | -4 = 0 and | z - i | - | z + 5i |= 0 (a)

(d)

0

Putting z = x + iy, these equations become | x + iy |= 4 i.e. x 2 + y 2 = 16

...(1)

and | x + iy - i |=| x + iy + 5i | or x 2 + ( y - 1) 2 = x 2 + ( y + 5) 2

i.e

y= –2

...(2)

Laws of Motion \ z = 1 + cos q + i sin q

Hence the complex numbers z satisfying the given equations are z1 = (2 3 , - 2) and z 2 = (-2 3 , - 2) that is, z1 = 2 3 - 2i, z 2 = -2 3 - 2i 6.

If S(n ) = i n + i - n , where i =

-n n n Sol. We have, S( n ) = i + i = i +

=

( -1) + 1 n

in

- 1 and n is a positive

1 in

=

in

q 1 q q q \ arg z = = arg(z - 1) + 2i sin cos 2 2 2 2 2 Thus, arg(z - 1) = 2 arg z. Answer (a) Let z be a complex number having the argument = 2 cos 2

9.

p and satisfying the equation | z – 3i | = 3. 2

6 is z Sol. From geometrical representation we know that the equation | z – 3i| = 3 represents a circle with centre C (3i) i.e., the point (0, 3) and radius = 3 Clearly, the circle touches the x - axis at origin O. Let a point P on the circle is the image of z, which has argument q . hence ÐOAP = q By property of circle ÐOAP = q .

Find the value of cot q -

, n = 1, 2, 3, 4,...

\ Total number of distinct values of S(n ) is 3. Answer (c) If z1, z2 , z3, z4 are represented by the vertices of a rhombus taken in the anticlock wise order, then

(a) z1 + z 2 = z 3 + z 4

qé q qù q cos + i sin ú = 2 cos .e iq / 2 ê 2ë 2 2û 2

q, 0 < q <

i 2n + 1

\ values of S(n) are 0, –2, 0, 2, 0, –2, ....

7.

= 2 cos

Answer (b, c)

integer, then the total number of distinct values of S(n ) is (a) 1 (b) 2 (c) 3 (d) 4

23

(b) z1 - z 2 + z 3 - z 4 = 0

z2 - z4 p z1 - z 2 p = (d) amp z z = 2 z1 - z 3 2 3- 4 Sol. Since diagonals of a rhombus bisect each other

(c) amp

z1 + z 3 z 2 + z 4 = = z 0 (say) 2 2 Þ z1 - z 2 + z 3 - z 4 = 0 Also, since diagonals of a rhombus are at right angles \

z2

z1 O

z0

z3

\ amp

z4

z + z4 z2 - 2 p 2 Þ amp = z1 + z 3 2 z1 2

8.

z 2 - z0 p = z1 - z 0 2

Þ amp

z2 - z4 p = z1 - z 3 2

Answers (b), (c) If z in any complex number satisfying | z - 1 |= 1 , then which of the following is correct ? (a) arg(z - 1) = 2 arg z (b) 2 arg( z) =

2 arg( z 2 - z) 3

(c) arg(z - 1) = arg(z + 1) (d) arg z = 2 arg(z + 1) Sol. Since | z - 1 |= 1

\ z - 1 = eiq , where arg | z - 1 |= q

We have | z | = OP = OA sin q = 6 sin q \ z = 6 sin q (cos q + i sin q ) Þ

6 1 cos q - i sin q = = = cot q - i sin q z sin q (cos q + i sin q)

6 = i z 10. Find the real part of tan (a + ib) Sol. We have

\ cot q -

sin( a + i b ) 2 sin( a + ib ) cos( a - ib ) tan (a + ib) = cos( a + ib ) = 2 cos( a + ib ) cos( a - ib ) [Multiplying numerator and denominator by 2 cos (a – ib)]

æ e 2b - e -2b ö ÷ sin 2a + iç ÷ ç 2 sin 2a + sin(2ib) ø è = cos 2a + cos( 2ib) = æ e 2b + e - 2b ö ÷ cos 2a + ç ÷ ç 2 ø è

=

2 sin 2a + i(e 2b - e -2b ) 2 cos 2a + (e 2b + e - 2b )

é ù ei(2ib) - e-i(2ib) (Euler's rotation) ú ê Note that sin2ib = 2i ê -2b 2b ú ìï e2b - e-2b üï ê e -e ú = ií ý . Similarly for cos (2ib).ú ê= 2i 2 îï þï ë û

24

Ph y si cs

2 sin 2 a \ Real part of tan (a + ib) is 2 cos 2 a + ( e 2b + e - 2b )

1 11. If |z| ³ 3, then find the least value of z + z

Sol. : Let z = r (cos q + i sin q ). Then | z | = r ³ 3

1 = r (cos q + i sin q) + (cos q - i sin q) r

=

æ 1ö æ 1ö ç r + ÷ cos q + i ç r - ÷ sin q r rø è ø è

Now a being real implies a 2 is +ve and hence we

2

Þ-

2

1 1 1 + + = 2w 2 a+w b+w c+w

2

and

p ) 2

\ for a given value of r

2

æ 1ö - 2 = çr - ÷ 2 rø è r

2

+

1

1ö æ \ ( t ) least = ç r - ÷ rø è

Now r ³ 3 Þ t least ³ 3 - 1 = 8 . 3 3 1 8 = z 3

12. Find the set of exhaustive values of real number a for which the equation z + a | z - 1 | +2i = 0 has a solution Sol. Let z = x + iy We have, z + a | z - 1 | +2i = 0 Þ x + i( y + 2) + a ( x - 1) 2 + y 2 = 0

Equating real and imaginary parts

2

+

1 c + w2

1 1 1 + + =2 a +1 b +1 c +1 (a) 1 (b) 2

Sol. Since w 2 =

1

\ Least Value of t = z +

1

a+w b+w Then show that

æ 2 1 ö = çç r + 2 ÷÷ + 2 cos 2 q r ø è

t2 is least if cos2 q = – 1 (occurs when q =

é 5 ùé 5ù 5 is –ve or êa + 2 ú êa - 2 ú is -ve êë 4 ûú êë ûú

5 5 £a£ 2 2 13. If w is complex cube root of unity and a, b, c are three real numbers such that

2

æ 1ö æ 1ö 2 2 = ç r + ÷ cos q + ç r - ÷ sin q r r è ø è ø

2 (t2)least = r +

5ö æ Þ 4a 2 ç a 2 - ÷ £ 0 4ø è

2 conclude that a -

1 Let t = z + z

t2

Þ 4a 4 + 20a 2 (1 - a 2 ) ³ 0 Þ -4a 4 + 5a 2 ³ 0

1

(c) 3

(d) None

1 1 and w = 2 the given relation may be w w 1 1 1 2 + + = a +w b+w c+w w

rewritten as and

= 2w

+

1

+

1

=

2

a+w b+w c+w w2 Clearly w and w2 are the roots of 2

2

2

1 1 1 2 + + = a+x b+x c+x x

...(1)

or

(b + x )(c + x) + (a + x )(c + x ) + (a + x )(b + x ) 2 = (a + x )(b + x)(c + x ) x

or

x[3x 2 + 2(a + b + c)x + bc + ca + ab]

= 2 [abc + (bc + ca + ab)x + (a + b + c) x 2 + x 3 ]

y + 2 = 0 \ y = -2 and x + a ( x - 1) 2 + 4 = 0

Þ x 3 - (bc + ca + ab)x - 2abc = 0 Now if a is the third root of this equation then sum of the

\ x 2 = a 2 ( x 2 - 2x + 5) or (1 - a 2 )x 2 + 2a 2 x - 5a 2 = 0

roots, a + w + w2 = 0 Þ a = 1 Hence, 1 is the root of equation (1) we get

Since x is real, \ D = B2 - 4AC ³ 0 (For a 2 = 1, the equation already has solution)

1 1 1 + + =2 a +1 b +1 c +1

Answer (b)

Laws of Motion

Fill in the Blanks 1. 2. 3.

4. 5.

p If | z | = 2 and arg (z) = then z = ............. 4 The greatest and the least absolute value of z + 1 where | z + 4 | £ 3 is ............. and ............. In the Argand plane, the vector z = 4 – 3i is turned in the clockwise sense through 180° and stretched 3 times. The complex number represented by the new vector is .............

The value of | a | , where a is a non-real cube root of unity is ............. For any two complex numbers | az1– bz2 |2 + | bz1 + az2 |2 ............. a + ib , then (x2 + y2)2 = ............. c + id

6.

If x + iy =

7.

The real value of ‘a’ for which 3i3 – 2ai2 + (1 – a)i + 5 is real is .............

8.

If zn = cos ç

æ

ö p æ ö p ÷ + sin ç ÷, è (2n + 1) (2 n + 3) ø è (2n + 1) (2 n + 3) ø

then lim (z1 z2 ...... z3) is equal to ............. n ®¥

The real value of ‘a’ for which 3i3 – 2ai2 + (1 – a) i + 5 is real is ............. 10. If (2 + i) (2 + 2i) (2 + 3i) .... (2 + ni) = x + iy then 4. 8. 13 ....... (4 + n2) = ............. 9.

11.

The value of (- -1)4 n-3 , where n Î N, is ..... True or False

12. Multiplication of a non-zero complex number by i rotates it through a right angle in the anti-clockwise direction. 13. The complex number cos q + i sin q can be zero for some q. 14. If a complex number coincides with its conjugate, then the number must lie on imaginary axis. 15. The argument of the complex number

z = (1 + i 3) (1 + i) (cos q + i sin q) is 7p + q . 12 16. The points representing the complex number z for which | z + 1 | < | z – 1 | lies in the interior of a circle. 17. If three complex numbers z1, z2 and z3 are in A.P., then they lie on a circle in the complex plane. 18. If n is a positive integer, then the value of in + (i)n+1 + (i)n+2 + (i)n+3 is 0.

25

19. The points having affixes z1, z2, z3 form an equilateral triangle iff

1 1 1 + + =0. z1 - z2 z2 - z3 z3 - z1

20. Multiplication of a non-zero complex number by – i rotates the vector through a right angle in counter clockwise sense. 21. If z1, z2, z3 respectively are affixes of points A, B, C then, æ z -z ö ÐCBA = ç 1 2 ÷ è z3 - z2 ø 22. If three complex numbers are in A.P., then they lie on a circle in the complex plane. 23. The trigonometric form of the complex number

z = 1 + i tan a where – p < a < p, a ¹ ± p / 2, is

1 cos α

(cos a + i sin a). 24. The points representing the complex numbers z for which | z + 1 | < | z – i | lie on a circle. 25. If a complex number coincides with its conjugate, then the number must lie on the imaginary axis. 26. The fourth roots of – 1, if plotted, would lie at the vertices of a square. 27. If the complex numbers z1, z2, z3 represent the vertices of an equilateral triangle such that | z1 | = | z2 | = | z3 |, then z1 + z2 + z3 ¹ 0 28. If in the Argand plane z1, z2, z3, z4 are four points such that | z1 | = | z2 | = | z3 | = | z4 |, then the four points are the vertices of a square. Short Answer Questions 29. If 4x + i (3x – y) = 3 + i (– 6), where x and y are real numbers then find the values of x and y. 30. Express the following in the form x + iy (i)

5 + 2i 1- 2i

(iii) i–35 + i– 40 31. If

3

1 ö æ (ii) ç - 2 - i ÷ 3 ø è (iv)

2-i (1 + 2i ) 2

x + iy = ± (a + ib) then show that

- x - iy = ± (b - ia )

æ z - 8i ö 32. If Re ç ÷ = 0 then show that z lies on the curve è z +6 ø x2 + y2 + 6x – 8y = 0

33. Find the conjugate of

(3 - 2i ) (2 + 3i ) (1 + 2i ) (2 - i )

26

Ph y si cs

34.

Evaluate (1 + i)6 + (1 – i)3

35.

Let z1 and z2 be two complex numbers such that | z1 + z2 | = | z1 | + | z2 |. Then show that arg (z1) – arg (z2) = 0 If z1, z2, z3 are complex numbers such that | z1 | = | z2 | = | z3|

36.

=

37.

1 1 1 + + z1 z 2 z 3

(i) sin 2a + sin 2b + sin 2g = 0 (ii) cos 3a + cos 3b + cos 3g = 3 cos (a + b + g) 43.

39.

40.

Convert the complex number z =

i -1 in the p p cos + i sin 3 3

n®¥

44. 45.

If | z + 1 | = z + 2 (1 + i), then find z. If arg (z – 1) = arg (z + 3i) then find (x – 1) : y, where z = x + iy

46.

Show that

u v If (x + iy)3 = u + iv then show that + = 4( x 2 - y 2 ) x y

47.

and radius. If | z1 | = | z2 | = ....... = | zn | = 1 then show that

42.

1.

2.

3.

4.

z -2 = 2 represents a circle. Find its centre z -3

| z1 + z2 + .....+ zn | =

1 1 1 + + ...... + z1 z2 zn

If a complex number z lies in the interior or on the boundary of a circle of radius 3 unit and centre (– 4, 0) then find the greatest and least values of | z + 1|. Find the value of 2x 4 + 5x 3 + 7x 2 – x + 41, when

48.

Find the complex number satisfying the equation

x = -2 - 3i

49.

z + 2 | z + 1| + i = 0 If z and w are two complex numbers such that | zw | = 1

Long Answer Questions 41.

ö ÷ , n Î N then prove that ø

lim (x1 x2 ......... xn) = – 1

= 1, then show that | z1 + z2 + z3 | = 1

polar form. 38.

æ p ö æ p If xn = cos ç ÷ + i sin ç n è2 ø è 2n

and arg (z) – arg (w) =

Find the common roots of the equations z3 + 2z2 + 2z + 1 = 0 and z1985 + z100 + 1 = 0 If cos a + cos b + cos g = sin a + sin b + sin g = 0 then prove that

z and w are two nonzero complex numbers such that | z | = |w| and Arg z + Arg w = p then z equals [AIEEE 2002] (a) w (b) – w (c) w (d) – w If |z – 4| < |z – 2| then its solution is given by (a) Re(z)>0 (b) Re(z)3 (d) Re(z)>2 The locus of the centre of a circle which touches the circle | z – z1 | = a and | z – z2 | = b externally (z, z1 & z2 are complex numbers) will be [AIEEE 2002] (a) an ellipse (b) a hyperbola (c) a circle (d) none of these Let z1 and z2 be two roots of the equation z 2 + az + b = 0 , z being complex. Further , assume that the origin, z1 and z2 form an equilateral triangle. Then (a)

a 2 = 4b

(c) a 2 = 2b

(b) a 2 = b (d) a 2 = 3b

50.

If

p , then show that z w = – i 2

z -1 is a purely imaginary number then show that z lies z +1

on a circle. Find the centre and radius of the circle

5.

If z and w are two non - zero complex numbers such that p zw = 1 and Arg ( z ) - Arg (w ) = , then zw is equal to 2 [AIEEE 2003] (a) –i (b) 1 (c) –1 (d) i x

6.

7.

æ 1+ i ö If ç [AIEEE 2003] ÷ = 1 then è 1- i ø (a) x = 2n + 1 , where n is any positive integer (b) x = 4n , where n is any positive integer (c) x = 2n , where n is any positive integer (d) x = 4n + 1 , where n is any positive integer..

Let z and w be complex numbers such that z + i w = 0 and arg zw = p. Then arg z equals [AIEEE 2004] (a)

5p 4

(b)

p 2

(c)

3p 4

(d)

p 4

[AIEEE 2003]

Laws of Motion 1

8.

9.

æx yö If z = x - i y and z 3 = p + iq , then çç + ÷÷ (p 2 + q 2 ) èp qø is equal to [AIEEE 2004] (a) –2 (b) –1 (c) 2 (d) 1 2 2 If | z - 1 |=| z | +1, then z lies on [AIEEE 2004] (a) an ellipse (b) the imaginary axis (c) a circle (d) the real axis

10. If the cube roots of unity are 1, w , w2 then the roots of the equation ( x – 1) 3 + 8 = 0, are

11.

–1, – 1, – 1

(d) – 1, 1 + 2 w , 1 + 2 w

(a)

17.

18.

2

that | z1 + z 2 | = | z1 | + | z 2 | , then argg z1 – argg z 2 is equal to [AIEEE 2005] p 2

(b) – p

(c) 0 12. If w =

(d) z

å

2

2

2

+ . . . . . . . . . +

(b)

(c)

b Î (1, ¥)

(d) b Î (0,1)

0ù ú , then H70 is equal to – ωû

(a) 0 (c) H2

(b) –H (d) H

2 If z ¹ 1 and z is real, then the point represented by the

22.

complex number z lies : [AIEEE 2012] (a) either on the real axis or on a circle passing through the origin. (b) on a circle with centre at the origin (c) either on the real axis or on a circle not passing through the origin. (d) on the imaginary axis. If z is a complex number of unit modulus and

z -1

æ 1+ z ö argument q, then arg ç equals: è 1 + z ÷ø [AIEEE 2006]

(a) 18 (b) 54 (c) 6 (d) 12 15. If | z + 4 | £ 3, then the maximum value of | z + 1 | is [AIEEE 2007] (a) 6 (b) 0 (c) 4 (d) 10

[AIEEE 2011]

21.

2

æ 6 1 ö ç z + 6 ÷ is z ø è

b Î (-1, 0)

éω H = ê ë0

[AIEEE 2006] (a) i (b) 1 (c) – 1 (d) – i 14. If z2 + z + 1 = 0, where z is complex number, then the value of 1ö æ 2 1 ö æ 3 1 ö æ çz + z ÷ + çz + 2 ÷ + çz + 3 ÷ è ø è z ø è z ø

b =1

(a)

20.

(b) a circle (d) a parabola

2kp 2kp ö æ 13. The value of + i cos ÷ is ç sin 11 11 ø k =1 è

–1 1 (d) i +1 i –1 The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals [AIEEE 2010] (a) 1 (b) 2 (c) ¥ (d) 0 Let a, b be real and z be a complex number. If z2 + az + b = 0 has two distinct roots on the line Re z =1, then it is necessary that : [AIEEE 2011]

Then (A, B) equals [AIEEE 2011] (a) (1, 1) (b) (1, 0) (c) (–1, 1) (d) (0, 1) If w ¹ 1is the complex cube root of unity and matrix

-p 2

10

1 i +1

If w( ¹ 1) is a cube root of unity, and (1 + w )7 = A + Bw.

[AIEEE 2005] (a) an ellipse (c) a straight line

(b)

19.

and | w | = 1, then z lies on

1 z- i 3

–1 i –1

(c)

If z1 and z 2 are two non- zero complex numbers such

(a)

1 then that i –1 [AIEEE 2008]

The conjugate of a complex number is complex number is

[AIEEE 2005]

(a) –1, –1 + 2 w , – 1 – 2 w 2 (b) (c) – 1, 1 – 2 w , 1 – 2 w 2

16.

27

p –q 2 (c) q (d) p – q If the cube roots of unity are 1, w, w2, then the roots of the equation (x – 1)3 + 8 = 0 are [IIT 1979] 2 (a) – 1, 1 + 2w, 1 + 2w (b) – 1, 1 – 2w, 1 – 2w2 (c) – 1, – 1, – 1 (d) None of these

(a) –q

23.

[JEE M 2013]

(b)

28 24.

Ph y si cs

The smallest positive integer n for which

[IIT 1980]

34.

(1 + i )n1 + (1 + i 3 )n1 + (1 + i 5 ) n2 + (1 + i 7 ) n2 , where

n

æ 1+ i ö çè ÷ = 1 is 1- iø

25.

(a) (b) (c) (d)

z – 5i = 1 lie on z + 5i

[IIT 1981]

28.

29.

30.

31.

32.

36. [IIT 1982]

by [IIT 1982] (a) Re(z) ³ 0 (b) Re(z) < 0 (c) Re(z) > 0 (d) none of these If z = x + iy and w = (1–iz)/ (z–1) then |w| = 1 implies that, in the complex plane, [IIT 1983] (a) z lies on the imaginary axis (b) z lies on the real axis (c) z lies on the unit circle (d) None of these The points z1, z2, z3, z4 in the complex plane are the vertices of a parallelogram taken in order if and only if [IIT 1983] (a) z1 + z4 = z2 + z3 (b) z1 + z3 = z2 + z4 (c) z1 + z2 = z3 + z4 (d) None of these If a, b, c and u, v, w are complex numbers representing the vertices of two triangles such that c = (1 – r) a + rb and w = (1 – r)u + rv, where r is a complex number, then the two triangles [IIT 1985] (a) have the same area (b) are similar (c) are congruent (d) none of these If w (¹ 1) is a cube root of unity and (1 + w)7 = A + Bw then A and B are respectively [IIT 1995S] (a) 0, 1 (b) 1, 1 (c) 1, 0 (d) – 1, 1 Let z and w be two non zero complex numbers such that | z | = | w | and Arg z + Argg w = p, then z equals (a) w (b) - w [IIT 1995S]

33.

w

334

æ 1 i 3ö +3 ç- + ÷ 2 ø è 2

365

[IIT 1999] (b) – 1 + i 3

(d) - i 3 i 3 If arg(z) < 0, then arg (-z) - arg(z) = (b) - p (a) p (c)

37.

(d) - w

Let z and w be two complex numbers such that | z | £ 1, | w | £ 1 and | z + i w | = | z – i w | = 2 then z equals (a) 1 or i (b) i or – i [IIT 1995S] (c) 1 or – 1 (d) i or – 1

-

z1 = z2 = z3 =

38.

39.

[IIT 2000S]

p p (d) 2 2 If z1, z2 and z3 are complex numbers such that [IIT 2000S]

(c)

The inequality z – 4 < z – 2 represents the region given

(c)

If i =

(a) 1 - i 3

5

æ 3 iö æ 3 iö + ÷ +ç - ÷ , then If z = ç è 2 2ø è 2 2ø

æ 1 i 3ö - 1 , then 4 + 5 ç - 2 + 2 ÷ è ø

is equal to

(a) Re(z) = 0 (b) Im(z) = 0 (c) Re(z) > 0, Im (z) > 0 (d) Re(z) > 0, Im (z) < 0 27.

35.

the x-axis the straight line y = 5 a circle passing through the origin none of these 5

26.

i = – 1 is a real number if and only if [IIT 1996] (a) n1 = n2 +1 (b) n1 = n2 –1 (c) n1 = n2 (d) n1 > 0, n2 > 0

(a) n = 8 (b) n = 16 (c) n = 12 (d) none of these The complex numbers z = x + iy which satisfy the equation

For positive integers n1, n2 the value of the expression

1 1 1 + + = 1, then z1 z2 z3

z1 + z2 + z3

is (a) equal to 1 (b) less than 1 (c) greater than 3 (d) equal to 3 Let z1 and z2 be nth roots of unity which subtend a right angle at the origin. Then n must be of the form [IIT 2001S] (a) 4k + 1 (b) 4k +2 (c) 4k + 3 (d) 4k The complex numbers z 1, z 2 an d z 3 satisfying z 1- z 3 1 - i 3 = are the vertices of a triangle which is z 2 -z 3 2

40.

41.

(a) of area zero [IIT 2001S] (b) right-angled isosceles (c) equilateral (d) obtuse-angled isosceles For all complex numbers z1, z2 satisfying |z1|=12 and | z2-3-4i | = 5, the minimum value of |z1-z2| is[IIT 2002S] (a) 0 (b) 2 (c) 7 (d) 17 Let w = 1

1

1 - 1 - w2 1 w2

1 3 +i , then the value of the det. 2 2 1 w2 is w4

[IIT 2002]

(a) 3w

(b) 3w(w - 1)

(c) 3w2

(d) 3w(1 - w)

29

Laws of Motion 42. If z = 1 and w =

z -1 ( where z ¹ -1) , then Re( w ) is z +1

(a) 0

(c)

(b)

z 1 . z +1 z +12

(d)

-

1 z +1

2

[IIT 2003S]

From z1 the particle moves

2 units in the direction of the

p in 2 anticlockwise direction on a circle with centre at origin, to reach a point z2. The point z2 is given by [IIT 2008] (a) 6 + 7i (b) –7 + 6i (c) 7 + 6i (d) –6 + 7i

vector iˆ + ˆj and then it moves through an angle

2 z +1

49. A particle P starts from the point z0 = 1 + 2i, where i = -1 . It moves horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1.

2

43. If w (¹ 1) be a cube root of unity and (1 + w2)n = (1 + w4)n, then the least positive value of n is [IIT 2004S] (a) 2 (b) 3 (c) 5 (d) 6 44. The locus of z which lies in shaded region (excluding the boundaries) is best represented by [IIT 2005S]

50. Let z = cos q + i sin q. Then the value of

m =1

at q = 2° is ( -1 + 2 , 2 )

A (–1, 0)

B

C (1, 0)

( - 1 + 2 ,- 2 )

arg (z) < p 4 arg ( z) > – p 4

3 2

(d)

1 2

w - wz 46. If is purely real where w = a + ib, b ¹ 0 and z ¹ 1, 1- z then the set of the values of z is [IIT 2006] (a) {z : |z| = 1} (b) {z : z = z } (c) {z : z ¹ 1} (d) {z : |z| = 1, z ¹ 1} 47. A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is [IIT 2007] (a) 3eip/4 + 4i (b) (3 – 4i)eip/4 (c) (4 + 3i)eip/4 (d) (3 + 4i)eip/4 z lie on 48. If | z | = 1 and z ¹ ± 1, then all the values of 1- z 2 (a) a line not passing through the origin

(b) | z | = 2 (c) the x-axis (d) the y-axis

[IIT 2009]

(a)

1 sin 2°

(b)

1 3sin 2°

(c)

1 2 sin 2°

(d)

1 4 sin 2°

51. Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are

D

(a) z : |z + 1| > 2 and |arg (z+1)| < p/4 (b) z : |z – 1| > 2 and |arg (z–1)| < p/4 (c) z : |z + 1| < 2 and |arg (z+1)| < p/2 (d) z : |z – 1| < 2 and |arg (z+1)| < p/2 45. a, b, c are integers, not all simultaneously equal and w is cube root of unity (w ¹ 1), then minimum value of |a + bw + cw2| is [IIT 2005S] (a) 0 (b) 1 (c)

15

å Im( z 2m –1 )

[IIT 2007]

–3 – 3 the roots of the equation z z + z z = 350 is[IIT 2009]

(a) 48 (b) 32 (c) 40 (d) 80 52. Let z be a complex number such that the imaginary part of z is non-zero and a = z2 + z + 1 is real. Then a cannot take the value [IIT 2012] (a) –1

(b)

1 3

1 2

(d)

3 4

(c)

53. Let complex numbers a and

1 lie on circles (x – x0)2 a

+ (y – y0)2 = r2 and (x – x0)2 + (y – y0)2 = 4r 2. respectively. If z0 = x0 + iy0 satisfies the equation 2

2 z 0 = r 2 + 2, then a = (a)

(c)

1 2 1 7

[JEE Adv. 2013]

(b)

1 2

(d)

1 3

30

1.

2.

3.

Ph y si cs

The number of solutions of the equation z2 + | z |2 = 0 where z ÎC is (a) one (b) two (c) three (d) infinitely many If a and b are the roots of x2 – x + 1 = 0, then the equation whose roots are a100 and b100 are (a) x2 – x + 1 = 0 (b) x2 + x – 1 = 0 2 (c) x – x – 1 = 0 (d) x2 + x + 1 = 0 If z be a complex number satisfying z4 + z3 + 2z2 + z + 1 = 0 then |z| is equal to 1 3 (b) 2 4 (c) 1 (d) No unique value For a complex number z, the minimum value of | z | + | z – 2 | is (a) 1 (b) 2 (c) 3 (d) None The greatest and the least absolute value of z + 1,where | z + 4 | £ 3 are respectively (a) 6 and 0 (b) 10 and 6 (c) 4 and 3 (d) none

(a) 4.

5.

6.

If f(z) =

7-z 1- z2

, where z = 1+ 2i, then |f(z)| is equal to :

|z| (b) | z | (c) 2 | z | (d) None 2 Given that |z–a| = a where z is a point in the Argand plane,

(a) 7.

z - 2a = z (a) i tan (arg z) (c) tan (arg z)

then

8.

If x + iy =

3 then 4x – x2 – y2 reduces to : cos q + i sin q + 2

(a) 2 9.

(b) i cot (arg z) (d) none of these

(b) 3

(c) 4

(d) 5

If z1 = 3 + i 3 and z 2 = 3 + i , then the complex 50

æ z1 ö number çç ÷÷ lies in the : è z2 ø (a) first quadrant (c) third quadrant

(b) second quadrant (d) fourth quadrant

é æ a - ib öù 10. The value of Arg êi lnç ÷ú , where a and b are real ë è a + ib øû numbers, is

(a) 0 or p (c) not defined

p (b) 2 (d) none of these.

11. One of the values of i i is (a) e–p/2 (b) ep/2

(c) ep

(d) e–p

12. The solution of the equation z(z - 2i) = 2(2 + i) are (a) 3 + i, 3 – i (b) 1 + 3i, 1 – 3i (c) 1 + 3i, 1 – i (d) 1 – 3i, 1 + i 13. If the quadratic equation z2 + (a + ib) z + c + id = 0 where a, b, c, d are non-zero real number, has a real root then (a) abd = b2c + d2 (b) abc = bc2 + d2 2 2 (c) abd = bc + ad (d) none of these. 14. Let l Î R . If the origin and the non real roots of 2 z 2 + 2 z + l = 0 form the three vertices of an equilateral triangle in the argand plane. Then l is 2 (c) 2 (d) –1 3 15. Let OA.OB = 1 and let O, A, B, be three collinear points. If O and B represent the complex numbers 0 and z, then A represents :

(a) 1

(b)

1 1 (d) z 2 (b) (c) z z z 16. The area of the triangle on the Argand diagram formed by the complex number z, iz and z+iz is

(a)

| z |2 | z |2 (c) (d) None 3 2 17. If z, wz ane w z are the vertices of a triangle, then the area of the triangle will be (where w is cube root of unity) :

(a) |z|2

(a)

(b)

3 | z |2 2

(b)

3 3 | z |2 2

3 | z |2 (d) None of these 2 18. Let z lies on the circle centred at the origin. If area of the triangle whose vertices are z, wz and z+wz, where w is the (c)

cube root of unity is 4 3 sq. unit. Then radius of the circle is : (a) 1 unit (b) 2 units (c) 4 Units (d) none 19. If x = cos q + i sin q , y = cos f + isin f , z = cos y + isin y and

x y z = 1, + + y z x

then cos( f – y ) + cos( y – q ) + cos( q – f ) is (a)

3 2

(b) –

3 2

(c) 0

(d) 1

31

Laws of Motion 20. Let z = 1 – t + i

( t 2 + t + 2 ) , where t is a real parameter..

The locus of z in the Argand plane is (a) a hyperbola (b) an ellipse (c) a straight line (d) none of these. 21. The locus of a point in the Argand plane that moves satisfying the equation | z - 1 + i | - | z - 2 - i | = 3 : (a) (b) (c) (d)

3 is a circle with radius 3 and centre at z = 2 is an ellipse with its foci at 1 – i and 2 + i and major axis = 3 is a hyperbola with its foci at 1–i and 2 + i and its transverse axis = 3 none of the above

22. The locus represented by | z - 1 | = | z + i | is (a) a circle of radius 1 (b) an ellipse with foci at (1, 0) and (0, –1) (c) a straight line through the origin (d) a circle on the line joining (1, 0), (0, 1) as diameter. 23. The maximum value of |z| when z satisfies the condition z+

2 = 2 is z

28. Let z be a complex number of maximum modulus such that z +

1 = 1. Then z

(a) Im (z) = 0 (b) Re (z) = 0 (d) None of these (c) amp (z) = p 29. If a, b, g and a, b, c are complex numbers such that

a b c b g a + + = 1 + i and + + = 0, then the value of b c a b g a a2

+

b2

a2 b2 (a) 0

+

g2

is equal to c2 (b) – 1

(c) 2i

30. If z = x + iy, z1 / 3 = a – ib, then where k is equal to (a) 1 (b) 2

(d) – 2i

y x – = k ( a2 – b2 ) b a

(c) 3

(d) 4 iA

31. If A, B, C are the angles of a triangle and e , e iB , e iC are in A.P. Then the triangle must be (a) right angled (b) isosceles (c) equilateral (d) None of these 32. The principle value of the arg (z) and | z | of the complex

(a)

3 -1

(b)

3 +1

æ 11p ö æ 11p ö ÷ + i sin ç number z = 1 + cos ç ÷ are respectively.. 9 è ø è 9 ø

(c)

3

(d)

2+ 3

(a)

24. For a complex number z, it is given that z +

1 = 2 . Which z

of the following is incorrect? (a) | z |£

2 +1

(c) | z | ³ 4

(b) | z |³ (d)

2 -1

2 -1 £

1 £ z

2 +1

25. Let Z and W be two complex numbers such that Z £ 1,

W £ 1 and Z+ i W = Z – i W = 2. Then Z equals (a) 1 or i (b) i or – i (c) 1 or – i 26. If w is imaginary cube root of unity, then

(d) i or – 1

pü ì sin í(w13 + w2 ) p + ý is equal to 4þ î

3 1 1 3 (b) – (c) (d) 2 2 2 2 27. If n is a positive integer grater than unity and z is a complex satisfying the equation z n = (z + 1) n , then (a) Re(z) < 0 (b) Re(z) > 0 (d) z lies on x = –

(b) –

7p æ 11p ö , –2cos ç ÷. 18 è 18 ø

(d) –

p æpö , – 2cos ç ÷ 9 è 18 ø

33. If z = ( l + 3) + i 5 – l2 then the locus of z is a (a) circle (b) sphere (c) straight line (d) None of these 34. Let z = log 2 (1 + i), then (z + z ) + i (z – z ) = ln 4 + p p – ln 4 (b) ln 4 ln 2 ln 4 - p p + ln 4 (d) (c) ln 4 ln 2 35. The number of 25th roots of unity which are also 15th roots of unity is (a) 15 (b) 5 (c) 20 (d) 10

(a)

36. The solution 2 2 x 4 = ( 3 – 1) + i ( 3 + 1) are,

(a) –

(c) Re(z) = 0

11p æpö , 2cos ç ÷ 8 è 18 ø 2p æ 7p ö , 2cos ç ÷ (c) 9 è 18 ø

1 2

5p 5p ö æ (a) ± ç cos + i sin ÷ 48 48 ø è

29p 29p ö (b) ± æç cos + i sin ÷ 48 48 ø è

æ

7p 7p ö + isin ÷ 48 48 ø

æ è

19p 19p ö – isin ÷ 48 48 ø

(c) ± ç cos è (d) ± ç cos

32

Ph y si cs

37. If z1 = z 2 = ........... z n = 1, then the value of z1 + z 2 + ........ z n –

(a) 0

1 z1

+ 1

(b) 1

1 1 is, + ........+ z2 zn

(c) – 1

(d) None

A B + = 1. B A Then the origin and two points A and B form a triangle which is (a) equilateral (b) obtuse angled triangle (c) right angled triangle (d) None of these

38. Let A and B be complex numbers such that

45. For x Î R, and x > 1, the value of æ x –i ö ÷÷ – p + 2 tan – 1 x, is i log çç x + i è ø (a) 0 (b) 1 (c) – 1

pö æ 46. If z = 1 + i tan aç - p < a < - ÷ , then polar form of the 2ø è complex number z is:

(a)

1 (cos a + i sin a ) cos a

(b)

1 [cos ( p + a ) + i sin( p + a ) - cos a

39. The three points z1 , z 2 , z 3 are connected by the relation a z1 + b z 2 + c z 3 = 0, z1 , z 2 , z 3 are complex numbers and a + b + c = 0 a, b, c Î R . Then the points are (a) collinear (b) not collinear (c) linearly dependent (d) linearly independent 40. If z1 = a + ib and z 2 = c + id are complex numbers such that z1 = z 2 = 1 and Re (z1z 2 ) = 0 then the pair of complex numbers w 1 = a + ic and w 2 = b + id satisfy

w1 = 1

(a)

w2 = 1

(b)

(d) none of these (c) Re ( w1 w 2 ) = 0 41. Complex number z satisfies | z – a + ia | = 1 and has the least absolute value. Its absolute value is (a) a 2 – 1 (b) a 2 + 1 (c) 0 (d) 42. If ( w ¹ 1) is a cube root of unity, then 1 1+ w 1+ w

2

(a) 0

1+ w

1 + w2

1 1 + w2

1+ w 1

(b) 1

(c) – 4

a

(c)

(b)

x2 y2 + =1 8 10

(d)

(d) 2

x2 y2 + =1 9 40 x2 y2 + =1 3 25

44. If a , b are the roots of the equation t 2 – 2t + 2 = 0, a value of x Î R so that

sin nq ( x + a) n – (x + b) n = is a –b sin n q

(a) cot q – 1

(b) cot q + 1

(c) cos q

(d) tan q + 1

æ 2(z - 1) ö ÷ ç 47. If sin –1 ç 2 ÷ is defined for some z, where z is non( 1 + i ) ø è real, then: (a) Re (z) = 1, –1 £ Im (z) £ 1 (b) Re (z) = 1, Im (z) = 2 (c) Re (z) = 1, – ¥