Mathematics

Complex Numbers

Session

Session Objectives

Session Objective

1. Polar form of a complex number

2. Euler form of a complex number

3. Representation of z1+z2, z1-z2

4. Representation of z1.z2, z1/z2

5. De-Moivre theorem

6. Cube roots of unity with properties

7. Nth root of unity with properties

Representation of complex number in Polar or Trigonometric form

z = x + iy

z(x,y)

X

Y

O x

y

2 2x y

2 2

2 2 2 2

x yz x y i

x y x y

This you have learnt in the first

session

Representation of complex number in Polar or Trigonometric form

z = r (cos + i sin )

where r | z | and arg z

Examples: 1 = cos0 + isin0

-1 = cos + i sin

i = cos /2 + i sin /2

-i = cos (-/2) + i sin (-/2)

z(x,y)

X

Y

O x = rcos

y =rsin

r

2 2x y r

2 2

x xcos

rx y

2 2

y y, sin

rx y

Eulers form of a complex number

z = x + iy

z = r (cos + i sin )i iz re where e cos isin

i 2 2| e | cos sin 1 i| z | | r || e | r

iz r cos isin re

Examples:i ii0 i 2 21 e , 1 e , i e , i e

Express 1 – i in polar form, and then in euler form

i42 cos isin 2e

4 4

1 i1 i 2 2 cos isin

4 42 2

Properties of eulers form ei

ie cos isin

i 2 2| e | cos sin 1

iargzz | z |e remember this

ia ibcosa isina cosb isinb e e i a be cos a b isin a b

ia ibe e cosa isina cosb isinb a b a b a b

2cos cos isin2 2 2

a b

i2a b

2cos e2

a biia ib 2a b

e e 2isin e2

Illustrative Problem

Solution:

i = cos(/2) + i sin(/2) = ei/2

ii = (ei/2)i = e-/2

The value of ii is ____

a) 2 b) e-/2

c) d) 2

iiNow find i

Ans : i

Illustrative Problem

Find the value of loge(-1).

Solution:

-1 = cos + i sin = ei

loge(-1) = logeei = i

General value: i(2n+1), nZ

As cos(2n+1) + isin(2n+1) = -1

Illustrative Problem

If z and w are two non zero complex numbers such that |zw| = 1, and Arg(z) – Arg(w) = /2, then is equal to

a) i b) –i

c) 1 d) –1

z w

Solution

iArg(z)By eulers form z | z | e iArg zw

zw | zw | e

| zw | | z | | w | | z | | w | | zw | 1(given)

Arg zw Arg z Arg w Arg z Arg w2

i2zw e i

Representation of z1+z2

z1 = x1 + iy1, z2 = x2 + iy2

z = z1 + z2 = x1 + x2 + i(y1 + y2)

O A

z1(x1,y1)

z2(x2,y2)z(x1+x2,y1+y2)

B

2 1Note OAz z Bz

Oz1 + z1z Oz

ie |z1| + |z2| |z1 + z2|

Representation of z1-z2

z1 = x1 + iy1, z2 = x2 + iy2

z = z1 - z2 = x1 - x2 + i(y1 - y2)

z2(x2,y2)

z1(x1,y1)

-z2(-x2,-y2)z(x1-x2,y1-y2)

O

Oz + z1z Oz1

ie |z1-z2| + |z2| |z1|

|z1-z2| ||z1| - |z2||

also |z1-z2| + |z1| |z2|

Representation of z1.z2

z1 = r1ei1, z2 = r2ei2

z = z1.z2 = r1r2ei(1+ 2)

1 2 1 2 1 2| z z | r r | z | | z |

1 2 1 2 1 2Arg z z Arg z Arg z

r1ei1

r2ei2

r1r2ei(1+ 2)

O x

Y

1

2

1+ 2

Representation of z1.ei and z1.e-i

z1 = r1ei1

z = z1. ei = r1ei(1+ )

1 1| z | r and Arg z

r1ei1

r1ei(1+ )

O x

Y

1

r1ei(1- )

What about z1e-i

Representation of z1/z2

z1 = r1ei1, z2 = r2ei2

z = z1/z2 = r1/r2ei(1- 2)

11 1

2 2 2

zz rz

z r z

11 2 1 2

2

zArg z Arg Arg z Arg z

z

r1ei1

r2ei2

r1/r2ei(1- 2)

1

2

1- 2

Illustrative Problem

If z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then

a) a2 = 3b b) a2 = 4b

c) a2 = b d) a2 = 2b

Solution

/3 z1

z2 = z1ei /3

i / 3 i0 i / 31 2 1 1 1Sum z z z z e z e e

i6

12z cos e a .......I6

i2 23

1Squaring I we get 3z e a

i2 31 2 1Pr oduct z z z e b

Hence a2 = 3b

De Moivre’s Theorem

1) n Z,

ncos isin cosn isinn

ncos isin cosn isinn

m ncosa isina cosb isinb cos ma nb isin ma nb

m

n

cosa isinacos ma nb isin ma nb

cosb isinb

De Moivre’s Theorem

2) n Q,

cos n + i sin n is one of the values of (cos + i sin )n

n p / q

p / q 1 / qcos isin cosp isinp

2k p 2k pcos isin ,k 1,2,..., q 1

q q

Particular case

1 / n 2k 2kcos isin cos isin ,k 0,1,2,...,n 1

n n

1/ qcos p 2k isin p 2k

Illustrative Problem

Solution:

72

5

31

4

2 2cos5 isin5 cos isin

7 7Simplify

2 2cos4 isin 4 cos isin

3 3

2 725 5 7

1 324 4 3

cos isin cos isin

cos isin cos isin

2 2 1 2cos isin

3cos isin

cos3 isin3

Illustrative Problem

Solution:

n

1n n 2n

Pr ove that : 1 i 1 i 2 cos4

1 i 2 cos isin4 4

1 i 2 cos isin4 4

n

n 2n n

1 i 2 cos isin4 4

n

n 2n n

1 i 2 cos isin4 4

n n

1n n 2 2n n

1 i 1 i 2 2cos 2 cos4 4

Cube roots of unity

1

33x 1 or x 1

3 2x 1 0or x 1 x x 1 0

1 i 3 1 i 3x 1, ,

2 2

2 2or x 1, , or 1, ,

2i2 31 i 3 1 3 2 2

i cos isin e2 2 2 3 3

2i

31 i 3 1 3 2 2i cos isin e

2 2 2 3 3

Find using (cos0 + isin0)1/3

2

1 i 3 1 i 3Note :

2 2

2

1 i 3 1 i 3and

2 2

Properties of cube roots of unity

21 0

2 31. . 1 2| | | | 1

2 22

1, ,

1,, 2 are the vertices of equilateral triangle and lie on unit circle |z| = 1

Why so?

1

2

O

2

3

4

3

Illustrative Problem

If is a complex number such that

2++1 = 0, then 31 is

a) 1 b) 0

c) 2 d)

Solution

2 1 0

21 i 3or

2

31 31let ,

312 31 2 62 2or ,

Nth roots of unity

1 1n nx 1 cos0 isin0

2k 2kcos isin ,k 0,1,...,n 1

n n

i0k 0, x 1 e 2i

n2 2k 1, x cos isin e

n n

4

i 2n4 4k 2, x cos isin e

n n.

.

2 n 1i n 1n

2 n 1 2 n 1k n 1, x cos isin e

n n

Properties of Nth roots of unity

n2 n 1 1

a) 1 ..... 01

n 1 n

n 12 n 1 2b) 1. . ...... 1

1 n is odd

1 n is even

c) Roots are in G.P

d) Roots are the vertices of n sided regular polygon lying on unit circle |z| = 1

2/n

4/n

n-1

1

23

Illustrative Problem

Solution:

Find fourth roots of unity.

1

4x 1 1

4cos0 isin 0 2k 2k

cos isin , k 0,1,2,34 4

k 0, x cos0 isin 0 1

k 1, x cos isin i2 2

k 2, x cos isin 1

3 3k 3, x cos isin i

2 2

-1 1

-i

i

Illustrative Problem

Solution:

3

4Find 1 in eulers form.

3

41 3

4cos isin

1

4cos3 isin3

2k 3 2k 3cos isin , k 0,1,2,3

4 4

3 5 7 9i i i i i

4 4 4 4 4e ,e ,e ,e e

Class Exercise

Class Exercise - 1

Express each of the following complex numbers in polar form and hence in eulers form.

(a) (b) –3 i 6 2 i

Solution

r | 6 2i |1. (a) 6 2 2 2

+ = 6

76

(– ) 6, – 2

1 2

tan66

7i

67 76 2i 2 2 cos isin 2 2 e

6 6

Solution Cont.

r 3i 0 9 3

13

tan0 2

–3i

32

3 3–3i 3 cos isin

2 2

b) –3i

3i

23e

Class Exercise - 2

Solution

| z z | | z | | z |1 2 1 2

| z | | z | | z z |1 2 1 2

z1 and z2 are in the same line

z1 and z2 have same argument or their difference is multiple of 2

arg (z1) – arg (z2) = 0 or 2n in general

(triangle inequality)

If z1 and z2 are non-zero complex numbers such that |z1 + z2| = |z1| + |z2| then arg(z1) – arg (z2) is equal to

(a) – (b) (c) 0 (d) 2

2

Class Exercise - 3

Solution

R (iZ)

Q (Z + iZ)

P (Z)

OX

Y

We have to find the area of PQR. Note that OPQR is a square as OP = |z| = |iz| = OR and all angles are 90°

Area of area of square OPQR 1

PQR2

1 12 2OP | z |2 2

Find the area of the triangle on the argand diagram formed by the complex numbers z, iz and z + iz.

Class Exercise - 4

If where

x and y are real, then the ordered pair (x, y) is given by ___.

503 i 3 253 (x iy),2 2

3 1a) ,

2 2

1 3c) ,

22

3 1d) ,

2 2

1 3b) ,

2 2

Solution

253 cos 8 isin 8

3 3

253 cos isin3 3

= 325 (x + iy)

1

x cos3 2

3y sin

3 2

50 503 3 3 i50i ( 3)2 2 2

503 i2532

503 i253

2 2

50253 cos isin

6 6

25 25253 cos isin3 3

253 cos 8 isin 8

3 3

Class Exercise - 5

Solution

Let x cos isin y cos isin

z cos isin

(cos cos cos ) i(sin sin sin )x + y + z =

If then prove that

cos cos cos 0 sin sin sin

cos3 cos3 cos3 3cos( )

sin3 sin3 sin3 3sin

= 0 + i0 = 0x3 + y3 + z3 – 3xyz

= (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = 0

x3 + y3 + z3 = 3xyz

Solution Cont.

x3 + y3 + z3 = 3xyz

3 3 3(cos isin ) (cos isin ) (cos isin )

3 cos( ) isin( )

comparing real part cos3 cos3 cos3 3cos( )

comparing imaginary part sin3 sin3 sin3 3sin

Class Exercise - 6

Solution

If and then

is equal to ___.

1

x 2cosx

1

y 2cos ,y

m nx yn my x

a)2cos b)2cos(m n )

c)2cos(m – n ) d) 0

1

x 2cosx 2x 2cos x 1 0

22cos 4cos 4x

2

2cos 2isin

2 cos isin

x cos isin

Take any one of the values say

Solution Cont.

x cos isin

y cos isinSimilarly

m n m nx y (cos isin ) (cos isin )n m n my x (cos isin ) (cos isin )

cos m – n isin m – n

cos n m isin n m

2cos m n

Class Exercise - 7

Solution

The value of the expression

2 21(2 )(2 ) ... (n 1)(n )(n ),

where is an imaginary cube root of unity is ___.

n 2(r 1)(r )(r )r 2

n 2 2 3(r 1)(r ( )r )r 2

n 2(r 1)(r r 1)r 2

n 3(r 1)r 2

n n3r 1r 2 r 2

n n3r – 1r 1 r 1

22n n 1n

4

Class Exercise - 8

Solution

If are the cube roots of p, p < 0, thenfor any x, y, z, is equal to

(a) 1 (b) (c) 2 (d) None of these

x y z

x y z

1

3x p , p 0

1 1 123 3 3p , p , p

123say p , ,

2x y z x y z2x y z x y z

2x y z2 2

2x y z

Class Exercise - 9

Solution:

The value of is

(a) –1 (b) 0

(c) i (d) –i

6 2k 2ksin – icos

7 7k 1

6 2k 2ksin – icos

7 7k 1

6 2k 2kcos isin 0

7 7k 0

6 2k 2k–i cos isin

7 7k 1...(i)

roots of x7 – 1 = 0 are

2k 2kcos isin ,

7 7 k = 0, 1, …, 6

Solution Cont.

6 2k 2kcos isin 0

7 7k 0

6 2k 2k1 cos isin 0

7 7k 1

6 2k 2ksin – icos (–i)(–1) i

7 7k 1

6 2k 2kcos isin –1

7 7k 1...(ii)

From (i) and (ii), we get

Class Exercise - 10

Solution:

If 1, are the roots of the equation xn – 1 = 0, then the argument of is

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

2 n 1, ,...,

2

2

n4

n

6

n

8

n

2

2n

2n

X

Y

O 1

As nth root of unity are the vertices of n sided regular polygon with each side making an angle of 2/n at the centre, 2 makes an angle of 4/n with x axis and hence, arg(2) = 4/n

Thank you