Complex Numbers

Complex Numbers

January 29, 2010

Abstract

These sheets are for reference for lectures taken on Complex numbers taught at IIT JEE level. Kindly provide your negative/positive negative/positive feedback on the content and their understanding.

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CONTENTS 

Contents 1

Origin & Motivation

3

2

Set of complex numbers and structure laid on it

3

2.0.1 2.0.2 2.0.3 2.0.4 2.0.5 2.0.6 3

Real part & Imaginary part . . . . . . . . . . . Geometry of complex numbers (Argand plane) Equality of complex numbers . . . . . . . . . . Additon or subtraction of complex numbers . Multiplication of complex numbers . . . . . . Division of complex numbers . . . . . . . . . .

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Three important properties of complex numbers

3.1 3.2 3.3

Conjugate of a complex number . . . . . . . . . . . 3.1.1 Conjugate properties . . . . . . . . . . . . . Modulus of a complex number . . . . . . . . . . . 3.2.1 Properties of Modulus . . . . . . . . . . . . Argument of a complex number . . . . . . . . . . . 3.3.1 Polar representation of a complex number .

4 4 5 5 5 5 7

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7 7 8 8 9 9

3

1

Origin & Motivation

We know solutions to equations like x2 1 = 0  but what about solution to equation x 2 +1 = 0. You can see that x 2 +1 = 0 x2 = 1. So that means we need to find a real number whose square is a negative number. Which cannot happen hence it was concluded that such a number has to be a totally new number outside set of real numbers.

−

2

⇒

−

Set of complex numbers and structure laid on it

We start with solving the above problem x2 = x =

−1√  ± −1

This new number was denoted using the letter ι   (read as iota) which is generally denoted by alphabet i  . Now we get something which is not real and hence the name imaginary numbers came into existence. Definition.   Imaginary numbers (Im) Im = ai/a So above solution to the equation x2 + 1 = 0 is x = ι and is a imaginary number Now we know that there are imaginary numbers other than real numbers which are worth studying to answer questions as possed above.

{

∈ }

 ±

Problem 1.   Solve the equation x2 i = 0 Solution :  Till here we know imaginary numbers are there for our rescue. So we solve the given problem

−

x2

= i x =

±

√ 

i

Now we see that we are not able to find the solution to this problem niether using real nor imaginary numbers. This motivates us to think that the search for new set is not still over and we come to the definition of complex numbers. Definition.   Complex numbers 1 C = a + ib/a, b R  and  i  = Notation : A complex number is usually represented by  z or ω or w

{

∈

√ −  }

Problem 2.   Find the value of  i k + ik+1 + ik+2 + ik+3 and i k + ik+2 (Note1 ) 1

Note here there are four consecutive powers of  i  and they are  1 , i, −1  & −i. Now note the location of these numbers in the argand plane. They lie at a distance of 1 from origin and on each positive, negative real & imaginary axis. Note that they are four in number and they are roots to the equation z4 − 1 = 0 . Prove this as an exercise.

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2

SET OF COMPLEX NUMBERS AND STRUCTURE LAID ON IT 

The value of any four consecutive powers of  i  is Zero. ik + ik+1 + ik+2 + ik+3

= = = =

ik (1 + i + i2 + i3 ) ik (1 + i 1 i) i(0) 0

− −

And ik + ik+2

= ik (1 + i2 ) = ik (1 1) = 0

−

Or rather we can think it in a different line, the first sum of four consecutive powers can be though as sum of four vectors that are along axes, and hence will cancel each other, the final outcome is zero doesnt depend on which of the four we start with. For the second problem, we start with either of four and add it to the just opposite of first selected. 2.0.1

Real part & Imaginary part

Given a complex number z  = a + ib we can say Re(z) =  a and is called the real part of  z  and similarly I m(z) =  b  is called the imaginary part of  z So a complex number can also be written as  z  = Re(z) + i Im(z) 2.0.2

Geometry of complex numbers (Argand plane)

Complex number a + ib can be thought of a coordinate pair (a, b) a + ib. So here we correspond the x-axis with Real axis and y-axis with Imaginary axis

≡

Cartesian coordinate system

Argand plane

x-axis y-axis Point (a, b)

Real axis Imaginary axis Complex number a + ib

Vector Space ˆi unit vector ˆ  j unit vector aˆi + b jˆ

5 2.0.3

Equality of complex numbers

Given two complex numbers z1 = a1  + ib1 and z2 = a2  + ib2 and z1 = z2 Re(z1 ) =  Re(z2 ) & I m(z1 ) =  I m(z2 ) i.e. a1  =  a 2 & b 1  = b 2

 ⇔

Problem 3.  Solve for x,y (x + iy) + (7 5i) = 9 + 4i x = 9 7 = 2 y = 4 + 5 = 9

⇒ ⇒

−

−

Problem 4.  True or false 2 + 3i  0 ; if b  0

z + Re(z) 2

z

Re(z) 2

; if Im(z)