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Mathematics. Session. Complex Numbers. Session Objectives. Session Objective. Polar form of a complex number Euler form of a complex number Representation of z 1 +z 2 , z 1 -z 2 Representation of z 1 .z 2 , z 1 /z 2 De-Moivre theorem Cube roots of unity with properties

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session
Session

Complex Numbers

session objective
Session Objective
  • Polar form of a complex number
  • Euler form of a complex number
  • Representation of z1+z2, z1-z2
  • Representation of z1.z2, z1/z2
  • De-Moivre theorem
  • Cube roots of unity with properties
  • Nth root of unity with properties
representation of complex number in polar or trigonometric form

z(x,y)

Y

y

O

X

x

Representation of complex number in Polar or Trigonometric form

z = x + iy

This you have learnt in the first session

representation of complex number in polar or trigonometric form1

z(x,y)

Y

r

y =rsin

O

X

x = rcos

Representation of complex number in Polar or Trigonometric form

z = r (cos  + i sin )

Examples: 1 = cos0 + isin0

-1 = cos  + i sin

i = cos /2 + i sin /2

-i = cos (-/2) + i sin (-/2)

eulers form of a complex number
Eulers form of a complex number

z = x + iy

z = r (cos  + i sin )

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

Examples:

illustrative problem

The value of ii is ____

  • 2 b) e-/2
  • c)  d) 2
Illustrative Problem

Solution:

i = cos(/2) + i sin(/2) = ei/2

ii = (ei/2)i = e-/2

illustrative problem1
Illustrative Problem

Find the value of loge(-1).

Solution:

-1 = cos  + i sin  = ei

loge(-1) = logeei = i

General value: i(2n+1), nZ

As cos(2n+1) + isin(2n+1) = -1

illustrative problem2

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

Illustrative Problem
representation of z 1 z 2

z(x1+x2,y1+y2)

z2(x2,y2)

z1(x1,y1)

B

O

A

Representation of z1+z2

z1 = x1 + iy1, z2 = x2 + iy2

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

Oz1 + z1z  Oz

ie |z1| + |z2|  |z1 + z2|

representation of z 1 z 21

z2(x2,y2)

z1(x1,y1)

O

z(x1-x2,y1-y2)

-z2(-x2,-y2)

Representation of z1-z2

z1 = x1 + iy1, z2 = x2 + iy2

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

Oz + z1z  Oz1

ie |z1-z2| + |z2|  |z1|

also |z1-z2| + |z1|  |z2|

 |z1-z2|  ||z1| - |z2||

representation of z 1 z 22

r1r2ei(1+ 2)

Y

1+2

r2ei2

r1ei1

2

1

O

x

Representation of z1.z2

z1 = r1ei1, z2 = r2ei2

z = z1.z2 = r1r2ei(1+ 2)

representation of z 1 e i and z 1 e i

Y

r1ei(1+ )

r1ei1

1

O

x

Representation of z1.ei and z1.e-i

z1 = r1ei1

z = z1. ei = r1ei(1+ )

What about z1e-i

r1ei(1- )

representation of z 1 z 23

2

r1ei1

1

r2ei2

r1/r2ei(1- 2)

1- 2

Representation of z1/z2

z1 = r1ei1, z2 = r2ei2

z = z1/z2 = r1/r2ei(1- 2)

illustrative problem3

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

  • a2 = 3b b) a2 = 4b
  • c) a2 = b d) a2 = 2b
Illustrative Problem
solution1

z2 = z1ei /3

z1

/3

Solution

Hence a2 = 3b

de moivre s theorem1
De Moivre’s Theorem

2) n  Q,

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

Particular case

cube roots of unity
Cube roots of unity

Find using (cos0 + isin0)1/3

properties of cube roots of unity

O

1

2

Properties of cube roots of unity

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

Why so?

illustrative problem6

If  is a complex number such that

2++1 = 0, then 31 is

  • 1 b) 0
  • c) 2 d) 
Illustrative Problem
properties of nth roots of unity

3

2

4/n

1

2/n

n-1

Properties of Nth roots of unity

c) Roots are in G.P

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

illustrative problem7

i

-1

1

-i

Illustrative Problem

Find fourth roots of unity.

Solution:

class exercise 1

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

(a) (b) –3 i

Class Exercise - 1

Solution

1. (a)

class exercise 2

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

(a) –p (b) (c) 0 (d)

Class Exercise - 2

Solution

(triangle inequality)

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

class exercise 3

Area of area of square OPQR

Class Exercise - 3

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

Solution

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

class exercise 4

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

Class Exercise - 4
solution3
Solution

= 325 (x + iy)

class exercise 5

If then prove that

Class Exercise - 5

Solution

x + y + z =

= 0 + i0 = 0

x3 + y3 + z3 – 3xyz

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

x3 + y3 + z3 = 3xyz

solution cont1
Solution Cont.

x3 + y3 + z3 = 3xyz

class exercise 6
Class Exercise - 6

If and then

is equal to ___.

d) 0

Solution

Take any one of the values say

class exercise 7

The value of the expression

wherew is an imaginary cube root of unity is ___.

Class Exercise - 7

Solution

class exercise 8

If are the cube roots of p, p < 0, then

for any x, y, z, is equal to

(a) 1 (b)  (c) 2 (d) None of these

Class Exercise - 8

Solution

class exercise 9

The value of is

  • –1 (b) 0
  • (c) i (d) –i

...(i)

k = 0, 1, …, 6

Class Exercise - 9

Solution:

roots of x7 – 1 = 0 are

solution cont3

...(ii)

Solution Cont.

From (i) and (ii), we get

class exercise 10

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

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

Class Exercise - 10

Solution:

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