Complex Numbers

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Complex Numbers

January 29, 2010

Abstract Abstract

These sheets are for reference for lectures taken on Complex numbers These sheets are for reference for lectures taken on Complex numbers taught at IIT JEE level.

taught at IIT JEE level. Kindly provide your negative/positiKindly provide your negative/positive feedbackve feedback on the content and their understanding.

on the content and their understanding.

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1 Origin & Motivation

We know solutions to equations like x2

−

1 = 0 but what about solution to equation x2_{+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 =

₋

1

x =

_±

√

₋

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

C₌

_{

_{a + ib/a, b}

_∈

R_{and i =}

√

₋

₁

_}

Notation : A complex number is usually represented by z or ω or w Problem 2. Find the value of ik

+ ik+1 _+ ik+2_+ ik+3 _and ik

+ 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 equationz4−1 = 0. Prove this as an exercise.

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The value of any four consecutive powers of i is Zero. ik

+ ik+1_+ ik+2_+ ik+3 _{= i}k

(1 + i + i2_+ i3₎ = ik (1 + i

₋

1

₋

i) = i(0) = 0 And ik + ik+2 _{= i}k (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 Vector Space

x-axis Real axis ˆi unit vector

y-axis Imaginary axis j unit vectorˆ Point (a, b) Complex number a + ib aˆi + bˆ j

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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 = a2 & b1 = b2

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 < 1

₋

i

We the question is absurd. Since there is no order deﬁned on set of C

numbers. So we cannot talk which complex number is greater. 2.0.4 Additon or subtraction of complex numbers Given complex numbers z1 = a1 + ib1 and z2 = a2 + ib2 then

z1

±

z2 = (a1

±

a2) + i(b1

±

b2)

Geometrically also we can see it as vector addition 2

2.0.5 Multiplication of complex numbers

Given a complex number z1 = a1 + ib1 and z2 = a2 + ib2 then

z1

·

z2 = (a1 + ib1)

·

(a2 + ib2)

= (a1a2

−

b1b2) + i(a1b2 + a2b1)

2.0.6 Division of complex numbers

Given a complex number z1 = a1 + ib1 and z2 = a2 + ib2 then

z1 z2 = z1 ¯z2

|

z2

|

2 = (a1 + ib1)(a2

−

ib2) a2 2 + b22

Problem 5. Now we revisit the problem discussed above, ﬁnd the roots of x2

−

i = 0

So we need to ﬁnd x =

_±

√

i Let the positive root be a + ib =

√

i (a + ib)2 ₌ _i (a2

−

b2_{) + i2ab} ₌ _i

⇒

a2

−

b2 _{= 0} _{and 2ab = 1}

⇒

a2= b2 and ab = 1 2 This solves to get us the solution as 1 + i

_√

2 and

−

1 + i

√

₂ . This can be seen more beautifully geometrically and easily guessed if we know multiplication is rotation of complex vectors.

2_{Remember : Complex numbers, Coordinate geometry and Vectors go hand in hand while}

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Example. Square root of a complex number a + ib Let the c + id =

√

a + ib (c2

₋

d2) + i(2cd) = a + ib c2

₋

d2 = a 2cd = b So we we can create c2+ d2 =



(c2

−

d2₎2_{+ 4c}2_d2 =



a2_+ b2 Therefore c =

_±



√

_a2_+ b2_+ a 2 d =

_±



√

_a2_+ b2

−

a 2 So we have

√

_{a + ib = c + id} =











±



√

_a2_+ b2_+ a 2 + i



√

a2_+ b2

−

a 2



; if b > 0

±



√

_a2_+ b2_+ a 2

−

i



√

a2_+ b2

−

a 2



; if b < 0 = =











±



|

z

|

+ Re(z)₂ + i



|

z

| −

Re(z) 2



; if Im(z) > 0

±



|

z

|

+ Re(z)₂

−

i



|

z

| −

Re(z) 2



; if Im(z) < 0 This can be easily written in one go as under

=

_±



|

z

|

+ Re(z) 2 + i S ignum(Im(z))



_|

z

_{| −}

Re(z) 2



where Signum(x) =



1 if x

≥

0

−

1 if x<0

We will touch the geometrically representation of the root once we get full throttle into the euler representation of complex number i.e. see multiplication as rotation in complex numbers.

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7

1. 9 + 40i 2. 50i

Using the above formula,

√

_{9 + 40i =}

±



41 + 9₂ + i



41

−

9 2



=

_±

(

√

25 + i

√

16) =

_±

(5 + 4i) And for the second problem,

√

_{50i =}

±

(5 + 5i)

3 Three important properties of complex

num-bers

3.1 Conjugate of a complex number

Given z = a + ib then conjugate of z is ¯

z = a

₋

ib

This can be looked upon in multiple ways one way is

•

Geometrically, z & ¯z are reﬂections in Real axis (x-axis)

•

Another way to look at conjugate of z is using transformation. Replacing every occurance of ι with

₋

ι.

Problem 7. Find the conjugate of the complex number z = 2 + i i(i

₋

3) +

2

₋

i i(3 + i) So conjugate of a complex number ¯z = z (try replacing i by

₋

i everywhere. 3.1.1 Conjugate properties 1. z1

±

z2 = z1

±

z2 2. z1z2 = z1

·

z2 3.



z1 z2



₌



z1 z2



4. z + z = 2Re(z) = 2Re(¯z) is always real (think in argand plane)

5. z

₋

z = 2i Im(z) =

₋

2i Im(¯z) is always imaginary (think in argand plane) Problem 8. For z1, z2

∈

C, value of I m(z1 ¯z2 + ¯z1z2) is

Note here z1z2 = ¯z1z2

⇒

Im(z1 ¯z2 + ¯z1z2) = Im(z1 ¯z2 + z1 ¯z2)

= Im(Re(z1 ¯z2))

= 0

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3.2 Modulus of a complex number

Modulus of a complex number z is denoted as

_|

z

_|

. It can be deﬁned in multiple similar ways as

•

For a complex number z = a + ib the modulus

_|

z

_|

=

√

a2_+ b2

•

OR Geometrically the distance3

of the point z in argand plane from origin. – Geometrically seeing is much easier using Pythogoras theorem.

Go-ing further we are goGo-ing to generalise the idea of modulus 3.2.1 Properties of Modulus

1. Triangle Inequality4

(a)

_|

z1 + z2

| ≤ |

z1

|

+

|

z2

|

equality is set z1 = kz2 where k > 0

(b)

_|

z1

−

z2

| ≥ | |

z1

| − |

z2

||

equality holds for z1 = kz2 where k < 0

2.

_|

z1z2

|

=

|

z1

||

z2

|

3.



z1 z2



=

|

z1

|

z2

|

4. Now comes the most important identity : z

_·

z¯ =

_|

z

_|

2

This property is important as it gives the movement to us from complex to real and real to complex.

Problem 10. If x + iy =



a + ib

c + id then prove that (x

2_+ y2_{) =} a 2_+ b2

c2_+ d2

3

Generalizing|z1−z2| =distance of complex number z1−z2 from origin. Geometrically

that can be seen as distance between z1 andz2

4_{Triangle inequality are used to ﬁnd the upper and lower bound of a complex number}

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3.3 Argument of a complex number 9

Problem 11. For a complex number z & w prove that

_|

z

_|

2_w

− |

w

_|

2_{z = z}

−

w iﬀ z = w or z ¯w = 1

This problem is a double implication i.e. if and only if . So ﬁrst lets prove if z = w or z ¯w = 1 then prove

_|

z

_|

2_w

− |

w

_|

2_z = z

−

w Proof : for z = w LHS =

_|

z

_|

2w

_{− |}

w

_|

2z =

_|

w

_|

2_w

− |

w

_|

w = 0 = w

₋

w = z

₋

w Or for z ¯w = 1 LHS =

_|

z

_|

2w

_{− |}

w

_|

2z = z(¯zw)

₋

w( ¯wz) = z(¯1)

₋

w(1) = z

₋

w Now we prove the converse if

_|

z

_|

2_w

− |

w

_|

2_{z = z}

−

w then z = w or z ¯w = 1 Proof :

|

z

_|

2w

_{− |}

w

_|

2z = z

₋

w w(

_|

z

_|

2_{+ 1) = z(}

|

w

_|

2_{+ 1)} w z =

|

w

_|

2_{+ 1}

|

z

_|

2_{+ 1}

that implies that w and z are scalar multiple of each other

∴ w = cz where c

∈

R+

Using this in the given we get

c

_|

z

_|

2z

₋

c2

_|

z

_|

2z = z

₋

cz c

_|

z

_|

2

−

c2

|

z

_|

2 _{= 1}

−

c c

_|

z

_|

2₍₁

−

c) = 1

₋

c c

_|

z

_|

2 = 1 or c = 1

⇒

w_z

|

z

_|

2 = 1 or z = w

⇒

w¯z = 1 or z = w

⇒

z ¯w = w ¯z = ¯1 = 1 or z = w Hence proved.

There can be a geometrical interpretation for the problem

3.3 Argument of a complex number

3.3.1 Polar representation of a complex number In argand plane, a complex number can be represented as

z = r(cos θ + i sin θ) = r(cisθ)

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where θ = arg(z) and r =

_|

z

_|

Deﬁnition. Argument5

of a complex number is deﬁned as the angle made by the complex number with the positive x

₋

axis. For a complex number z = a+ib we deﬁne it as arg(z) = tan−1



b

a



_{. Now any z = r cis(2π + θ) So there are} inﬁnite value of argument that are true for a given complex number hence we deﬁne principal argument denoted by Arg(z)

5_{argument of a complex number is a one to many mapping and hence needs to be deﬁned}

Complex Numbers

Complex Numbers

Complex Numbers

January 29, 2010

January 29, 2010

Contents

1 Origin & Motivation

−

⇒

−

2 Set of complex numbers and structure laid on

it

−

±

√

−

{

∈ }

±

−

±

√

{

∈

√

−

}

−

−

−

≡

⇔

−

⇒

−

⇒

−

±

±

±

·

·

−

|

|

−

−

±

√

√

−

⇒

−

⇒

√

−

√

√

−

−



−



±



√

±



√

−

√



















₋

_±

₋

_{

_{∈ }}

_±

_{

_∈

₋

_}

₋

₋

₋

_≡

₋

₋

₋

_±

_√

₋

₋

_±

_±

_±

_|

_{| −}

_±

_±