Complex Number

Introduction

A complex number is an extended version of real number in the form
x + iy; 𝑥∈ℝ
Euler (1707 – 1783) introduced the imaginary unit ‘i’ (read as iota) for √-1 with property
$i^2+ 1 = 0$
Therefore, imaginary unit i is the solution of an equation
$x^2+ 1 = 0$

Acomplex is written in the STANDARD form as z = x+iy where x and y are real numbers

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Note

The cartesian form of complex number can be written z=x+iy [standard form] or z=(x,y) [order pair form]
In z = (x,y,), x is real part denoted as Re (z), and y is imaginary part denoted as Im (z)
In z = (x,y), Re (z) = x and Im (z) =y
A complex number z = x+iy, is purely real if y = 0 i.e. Re (z) = 0, and purely imaginary if x = 0 i.e. Im (z) = 0.
A complex number z = x+iy, is zero if x = y = 0 i.e. Re (z) = Im (z) = 0

The introduction of complex numbers gives rise to the fundamental theorem of algebra. In the 16th century, Italian mathematician Gerolamo Cardano used complex numbers to find solutions to cubic equations.

Then Italian mathematician Rafael Bombelli developed the rules for addition, subtraction, multiplication, and division of complex numbers. A more abstract formalism for complex numbers was developed by the Irish mathematician William Rowan Hamilton.

Meaning of i

In the complex number system, i is called imaginary unit. Tthe value of i is (0, 1). Thus, we can write i=(0,1).

If we expand a complex number z=(x,y) as z=(1,0)x+(0,1)y then (1,0) is unit of real part denoted by 1 and (0,1) is unit of imaginary part and denoted by i. Here i is imaginary unit with
$i^2=-1$ .
A complex number is visually represented in Argand diagram. Here i is operator giving anticlockwise quarter turn such that $i^2=-1$ .

NOTE

We should NOT mean i as non-existence number nor a number that exist only in imagination
i is a number that denotes imaginary unit (0,1)
i is an anticlockwise quarter turn operator for (x,y), thus
$i^2=(0,1) (0,1) = i (0,1) =(-1,0) =-1 (1,0)=-1$

Positive powers of i

In general, for any integer k,
$i^{4k} = 1, i^{4k+1} = i, i^{4k+2} = -1, i^{4k+3} = -i$ .
For example:
$i^{23} = i^{4.5+3} = (i^4)5. i ^3= 1^5. (-i) = -i.$

The verification on the positive powers of i are as follows.
$i^2 = -1$
$i^3 = i2. I = (-1).i = - i$
$i^4 = (i^2)^2 = (-1)^2 = 1$
$i^5 = i^{4 + 1}= i^4. i = 1. i = i$
$i^6 = i^{4 + 2} = i^4. i^2= 1. 1 = 1$ and so on

Absolute Value

In a complex number
$z= x + iy$
The absolute value is Modulus, a non-negative real number denoted by
| z | and defined by
$|𝑧|=\sqrt{x^2+y^2}$
Geometrically,
|𝑧| is distance of z from origin

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Note
Due to order property, z₁ <z₂ is meaningless unless z₁ and z₂ both are real.
However,| z₁ | <| z₂ | means z₁is closer than z₂.
Distance between $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ is denoted by
$| z_1 -z_2 |$
and defined by
$| z_1 -z_2 |=\sqrt{( x_1 -x_2 )^2 +( y_1 -y_2 )^2 }$

NOTE:
$( x+iy )^2 =( x+iy )( x-iy )$

Conjugate

The conjugate of a complex number z=x+iy is denoted $\bar{z}$ and defined by
$\bar{z}=x-iy$
Geometrically, $\bar{z}$ is reflection of $z$ on real axis

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Properties of Conjugate numbers

If $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ then $\overline{z_1 +z_2 }=\bar{z_1} +\bar{z_2}$
Proof
$\overline{z_1 +z_2 }=\overline{( x_1 +iy_1 )+( x_2 +iy_2 )}$
or $\overline{z_1 +z_2 }=\overline{( x_1 +x_2 )+i( y_1 +y_2 )}$
or $\overline{z_1 +z_2 }=( x_1 +x_2 )-i( y_1 +y_2 )$
or $\overline{z_1 +z_2 }=( x_1 -iy_1 )+( x_2 -iy_2 )$
or $\overline{z_1 +z_2 }=\bar{z}_1 +\bar{z}_2$
The sum of a complex number z and its conjugate $\bar{z}$ is twice of its real part
Proof
If $z=x+iy$ be a complex number then $\bar{z}=x-iy$ , where
$z+\bar{z}=2x$
or $x=\frac{z+\bar{z}}{2}$
or $Rez=\frac{z+\bar{z}}{2}$
The difference of z and its conjugate $\bar{z}$ is twice of its imaginary part
Proof
If z=x+iy be a complex number then $\bar{z}=x-iy$ , where
$z-\bar{z}=2iy$
or $y=\frac{z-\bar{z}}{2i}$
or $Imz=\frac{z-\bar{z}}{2i}$
The real and imaginary parts of a complex number can be extracted using its conjugate

Theorem: An important property

Let z=x+iy be a complex number then $z.\bar{z}=|z|^2$

Proof
Given z=x+iy be a complex number, then $\bar{z}=xiy$ .
Now,
$z.\bar{z}=( x+iy )( x-iy )$
or $z.\bar{z}=( x )^2 -( iy )^2$
or $z.\bar{z}=x^2 +y^2$ (i)
Also
$| z |=\sqrt{x^2 +y^2}$
or $|z|^2 =x^2 +y^2$ (ii)
From (i) and (ii), we get
$z.\bar{z}=|z|^2$

Question

Theorem: Triangle inequality: $| z_1 +z_2 |\le | z_1 |+| z_2 |$

Proof>br> We know that
$| z_1 +z_2 |^2 =( z_1 +z_2 )( \overline{z_1 +z_2 } )$
or $| z_1 +z_2 |^2 =( z_1 +z_2 )( \bar{z}_1 +\bar{z}_2 )$
or $| z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +z_2 \bar{z}_1 +z_2 \bar{z}_2 )$
or $| z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +\bar{\bar{z}}_2 \bar{z}_1 +z_2 \bar{z}_2 )$
or $| z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +\overline{z_1 \bar{z}_2 }+z_2 \bar{z}_2 )$
or $| z_1 +z_2 |^2 =z_1 \bar{z}_1 +2Re( z_1 \bar{z}_2 )+z_2 \bar{z}_2$
or $| z_1 +z_2 |^2 \le z_1 \bar{z}_1 +2| z_1 \bar{z}_2 |+z_2 \bar{z}_2$
or $| z_1 +z_2 |^2 =|z_1|^2 +2| z_1 || \bar{z}_2 |+|z_2|^2$
or $| z_1 +z_2 |^2 =( | z_1 |+| z_2 | )^2$
or $| z_1 +z_2 |\le | z_1 |+| z_2 |$
This completes the proof.

Corollary

$| z_1 +z_2 |\ge | z_1 |-| z_2 |$

Proof
or $| z_1 |=| z_1 +z_2 +( -z_2 ) |$
or $\le | z_1 +z_2 |+| -z_2 |$
or $=| z_1 +z_2 |+| z_2 |$
Thus, $| z_1 +z_2 |\ge | z_1 |-| z_2 |$
$| z_1 -z_2 |\le | z_1 |+| z_2 |$

Proof
$| z_1 +z_2 | \le | z_1 |+| z_2 |$
Now, replacing $z_2$ by $-z_2$ we get
$| z_1 -z_2 |\le | z_1 |+| -z_2 |$
Thus, $| z_1 -z_2 |\le | z_1 |+| z_2 |$
$| z_1 -z_2 |\ge | z_1 |-| z_2 |$

Proof
$| z_1 +z_2 | \ge | z_1 |-| z_2 |$
Now, replacing $z_2$ by $-z_2$ we get
$| z_1 -z_2 |\ge | z_1 |-| -z_2 |$
Thus, $| z_1 -z_2 |\ge | z_1 |-| z_2 |$

Algebra of complex number

Complex plane looks like an ordinary two-dimensional plane of z=( x,y ), but z=( x,y ) is a single number losing order axioms. Fundamental operations on complex number are defined as below.

Equality
Two complex numbers $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ are equal if
$x_1 =x_2 \text{ and } y_1 =y_2$
Addition
Sum of two complex numbers $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ is defined as
$z_1 +z_2 =( x_1 +iy_1 )+( x_2 +iy_2 )=( x_1 +x_2 )+i( y_1 +y_2 )$
According to definition, $z_1 +z_2$ corresponds to resultant vector addition.
Multiplication
Product/multiplication of two complex numbers $z_1 =x_1 +iy_1$ and $z_2 =x_2 +iy_2$ is defined as
$z_1 .z_2 =( x_1 +iy_1 ).( x_1 +iy_2 )$
or $z_1 .z_2 =x_1 x_2 +ix_1 y_2 +ix_2 y_1 +{{i}^{2}}y_1 y_2$
or $z_1 .z_2 =( x_1 x_2 -y_1 y_2 )+i( x_1 y_2 +x_2 y_1 )$
The product is neither scalar nor the vector product of ordinary vector analysis. This departure is due to $i^2=-1$

Also, complex number C is a field. Thus, the complex number satisfy all Field axioms as below.

Closer: $\forall a,b \in G \implies a \ast b \in G$
Additive associative
$\forall a,b,c \in G \implies (a+ b) + c=a + (b + c)$
Additive identity
$\forall a \in G \implies a \ast e =e \ast a =a$
Additive identity (0,0)
Additive Inverse
$\forall a \in G \implies a \ast (a^{-1}) =(a^{-1}) \ast a =e$
Additive inverse of z=x+iy is -z=-x-iy
Additive commutative
$\forall a,b \in G \implies a \ast b =b \ast a$
Distributive
$\forall a,b,c \in G \implies (a + b)\bullet c =a c+bc$
Multiplicative associative
$\forall a,b,c \in G \implies (a. b) . c=a . (b . c)$
Multiplicative identity
$\forall a \in G \implies a \ast e =e \ast a =a$
Multiplicative identity (1,0)
Multiplicative inverse
$\forall a \in G \implies a \ast (a^{-1}) =(a^{-1}) \ast a =e$
Multiplicative inverse of non-zero complex number z=x+iy is
$z^{-1} =\frac{x}{x^2 +y^2} +i\frac{-y}{x^2 +y^2}$
Proof
Let z=x+iy be non-zero complex number and $z^{-1} =u+iv$ be its multiplicative inverse, then
$zz^{-1} =( 1,0 )$
or $( x+iy )( u+iv )=( 1,0 )$
or $( xu-yv )+i( xv+uy )=( 1,0 )$
Comparing real and imaginary parts separately, we get
ux-vy=1 and uy+vx=0
Solving for u and v , we get
$u=\frac{x}{x^2 +y^2}$ and $v=\frac{-y}{x^2 +y^2}$

Hence,
$z^{-1} =\frac{x}{x^2 +y^2} +i\frac{-y}{x^2 +y^2}$
Multiplicative commutative
$\forall a,b \in G \implies a \ast b =b \ast a$

Algebric Structure

Axioms	Structure	Example
1-2	Semi Group	Example
1-3	Monoid	Example
1-4	Group	Example
1-5	Abelian Group	Example
1-6	Ring	Example
1-7	Assocoative Ring	Example
1-8	Assocoative Ring with Unity	Example
1-9	Division Ring (Skew Field)	Example
1-10	Field	Example

Polar form

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Let $z=x+iy$ be a complex number with magnitude r and amplitude $\theta$ , then
$x=r\cos \theta$ and $y=r\sin \theta$
Hence,
$z=x+iy=r\cos \theta +ir \sin \theta$
or $z=r( \cos \theta +i\sin \theta )$
Thus, a complex number z=x+iy is represented by polar coordinate $(r,\theta)$ , as
$z=r \cos \theta +i\sin \theta$
Here,
r is the length of z , and $\theta$ is argument of z with
$r=| z |=\sqrt{x^2 +y^2}$ , and $\theta =argz=\tan^{-1}( \frac{y}{x} )$

Example

Find polar form of complex number z=-5+5i

Solution
Given that
z=-5+5i
or z=5( -1+i)
or $z=5\sqrt{2}( -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i )$
or $z=5\sqrt{2}( \cos \frac{3\pi }{4}+\sin \frac{3\pi }{4}i )$

The relevance of complex number in polar form is that multiplication and division are simpler with this form than the Cartesian form.

Let $z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 )$ and $z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 )$ be two complex numbers, then
$z_1 z_2 =r_1 r_2 (\cos (\theta_1 +\theta_2 )+i\sin (\theta_1 +\theta_2 ))$
Proof
Given that $z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 )$ and $z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 )$ . Thus
$z_1 z_2 = r_1 (\cos \theta_1 +i\sin \theta_1 ) \times r_2 (\cos \theta_2 +i\sin \theta_2 )$
or $z_1 z_2 = r_1 r_2 (\cos \theta_1 +i\sin \theta_1 )(\cos \theta_2 +i\sin \theta_2 )$
or $z_1 z_2 = r_1 r_2 [\cos \theta_1 (\cos \theta_2 +i\sin \theta_2 )+i\sin \theta_1 (\cos \theta_2 +i\sin \theta_2 )$
or $z_1 z_2 = r_1 r_2 [\cos \theta_1 \cos \theta_2 +i\cos \theta_1\sin \theta_2 +i\sin \theta_1\cos \theta_2 -\cos \theta_1\sin \theta_2 ]$
or $z_1 z_2 = r_1 r_2 [(\cos \theta_1 \cos \theta_2-\cos \theta_1\sin \theta_2 ) +i(\cos \theta_1\sin \theta_2 +\sin \theta_1\cos \theta_2) ]$
or $z_1 z_2 = r_1 r_2 [\cos (\theta_1 +\theta_2) +i \sin (\theta_1+ \theta_2)]$

Theorem

$arg( z_1 z_2 )=arg z_1 +argz_2$

Proof

Let $z_1 =r_1 ( \cos \theta_1 +i \sin \theta_1 )$ and $z_2 =r_2 ( \cos \theta_2 +i\sin \theta_2 )$ then
$z_1 .z_2=r_1 .r_2 [ \cos ( \theta_1 +\theta_2 )+i\sin ( \theta_1 +\theta_2 ) ]$
Thus,
$arg( z_1 z_2 )=argz_1 +argz_2$

Note:
Any complex number z has infinite arguments; all differ by multiple of $2\pi$ . The principal value is in the interval $[ -\pi ,\pi ]$

Some important property

Let $z_1 \text{ and } z_2$ be two complex number then $arg( z_1 .z_2 )=argz_1 +argz_2$
The argument of product of two complex number is sum of their arguments.
Proof
Let $z_1 =r_1 e^{ i\theta_1}$ and $z_2 =r_2 e^{ i\theta_2}$ then
$z_1 z_2 =( r_1 e^{ i\theta_1} )( r_2 e^{ i\theta_2} )$
or $z_1 z_2 =( r_1 r_2 )e^{i( \theta_1 +\theta_2 )}$
Hence,
$arg( z_1 .z_2 )=\theta_1 +\theta_2 =argz_1 +argz_2$
Let $z_1 \text{ and }z_2$ be two complex number then,
$arg( \frac{z_1 }{z_2 } )=argz_1 -argz_2$

The argument of quotient of two complex number is difference of their arguments.
Argument of complex number of the form $z=a+i.0, a > 0$ is 0
Argument of complex number of the form $z=a+i.0, a < 0$ is $\pi$
Argument of complex number of the form $z=0+i.b, b > 0$ is $\frac{\pi }{2}$
Argument of complex number of the form $z=0+i.b, b < 0$ is $-\frac{\pi }{2}$

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If $z =r( \cos \theta +i\sin \theta )$ then show that $z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )$

Question

Let $z=r( \cos \theta +i\sin \theta )$ be a complex number and its inverse is $z^{-1} =R( \cos \phi +i\sin \phi )$ then
$zz^{-1} =1$
or $r( \cos \theta +i\sin \theta )R( \cos \phi +i\sin \phi )=( 1,0 )$
or $rR\{ \cos ( \theta +\phi )+i\sin ( \theta +\phi ) \} =( 1,0 )$
or $rR\cos ( \theta +\phi )=1 \text{ and } rR\sin ( \theta +\phi )=0$
or $R=\frac{1}{r},\phi =-\theta$
Thus,
$z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )$

Exponenrial Form

Euler’s formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. It states that for any real number x:
$e^{i \theta}=\cos \theta+i\sin \theta$
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions.

When x = π, Euler’s formula evaluates to $e^{i\pi}+1=0$ , which is known as Euler’s identity.

Given Euler’s exponential form,
$e^{i\theta}=\cos \theta +i\sin \theta$
Thus, complex number $z=r( \cos \theta +i\sin \theta )$ is defined as
$z=re^{i\theta}$
The significance of exponential form of complex number is that we can easily compute conjugate and inverse.
For example,
$\bar{z}=re^{- i\theta}$ and $z^{-1} =\frac{1}{r} e^{- i\theta}$

Proof of the Formula

Function Method

Consider the function f(θ) given by
$f(\theta )=\frac {\cos \theta +i\sin \theta }{e^{i\theta }}$
or $f(\theta )= e^{- i\theta } ( \cos \theta +i\sin \theta)$
Differentiating gives by the product rule, we have
$f' (\theta )=0$
Thus, f(θ) is a constant.
Since f(0) = 1, then f(θ) = 1 for all θ, and thus
$1= e^{- i\theta } ( \cos \theta +i\sin \theta)$
or $e^{ i\theta }= \cos \theta +i\sin \theta$
This completes the proof.
Series Method

We know that
$e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$
$\cos x= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}-$
$\sin x= x-\frac{x^3}{3!} +\frac{x^5}{5!}-...$
Using power series expansion, we get
$e^{ i\theta }= \sum_{n=0}^\infty \frac{(i\theta)^n}{n!}$
or $e^{ i\theta }= 1+ \frac{(i\theta)^1}{1!}+ \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!}+ \frac{(i\theta)^4}{4!}+...$
or $e^{ i\theta }= 1+ i\theta- \frac{\theta^2}{2!} -i \frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+...$
or $e^{ i\theta }= \left ( 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}-...\right ) +i \left (\theta-\frac{\theta^3}{3!} +... \right )$
or $e^{ i\theta } = \cos \theta +i\sin \theta$
This completes the proof.

De-Moivre’s theorem

Let $z=r \cos \theta +i\sin \theta$ be a complex number then $z^n =r^n (\cos n\theta +i\sin n\theta )$ where n is a positive integer.
Proof

Case 1: n=1
Then
$z =r (\cos \theta +i\sin \theta )$
or $z^1 =r^1 (\cos 1.\theta +i\sin 1.\theta )$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=1
Case 2: n=2
$z^2 =z.z$
or $z^2 = r (\cos \theta +i\sin \theta ) \times r (\cos \theta +i\sin \theta )$
or $z^2 = r^2 (\cos \theta +i\sin \theta )(\cos \theta +i\sin \theta )$
or $z^2 = r^2 [\cos \theta (\cos \theta +i\sin \theta )+i\sin \theta (\cos \theta +i\sin \theta )$
or $z^2 = r^2 [\cos \theta \cos \theta +i\cos \theta\sin \theta +i\sin \theta\cos \theta -\cos \theta\sin \theta ]$
or $z^2 = r^2 [(\cos \theta \cos \theta-\sin \theta \sin \theta ) +i(\cos \theta\sin \theta +\sin \theta\cos \theta) ]$
or $z^2 = r^2 [\cos (\theta +\theta) +i \sin (\theta+ \theta)]$
or $z^2 = r^2 [\cos 2\theta +i \sin 2\theta]$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=2
Case 3: We assume the same formula is true for n = k, so we have
$(\cos\theta + i\sin\theta)^k = r^k(\cos(k\theta) + i\sin(k\theta))$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=k
Case 4: Now, we prove for n = k + 1,
$[r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos\theta + i\sin\theta)^k r (\cos\theta + i\sin\theta)$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos(k\theta) + i\sin(k\theta)) r(\cos\theta + i\sin\theta)$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} = r^{k+1}[(\cos(k\theta) \cos\theta-\sin(k\theta)\sin\theta )+i (\cos\theta\sin(k\theta) + \sin\theta\cos(k\theta))]$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k\theta+\theta)+i \sin(k\theta+\theta )]$
or $[r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k+1)\theta+i \sin(k+1)\theta]$
So,
or $z^n =r^n (\cos n\theta +i\sin n\theta )$ when n=k+1
Using case 1-case 4, for any number $n \in Z$ , we have
$[r(\cos\theta + i\sin\theta)]^n =r^n[\cos(n\theta)+i \sin(n\theta)]$

Example 1

Compute $(1+i)^6$
Solution
Since
$1+i= \sqrt{2} \left ( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right )$
or $1+i= \sqrt{2} \left ( \cos \frac{\pi }{4} +i\sin \frac{\pi }{4} \right )$
We get
$r=\sqrt{2}$ and $\theta=\frac{\pi }{4}$
Thus,
$( 1+i )^6 =\sqrt{2}^6 \left [ \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right ]^6$
or $( 1+i )^6 =\sqrt{2}^6[ \cos \left (6 \frac{\pi }{4} \right )+i\sin \left (6 \frac{\pi }{4} \right )]$
or $( 1+i )^6 =8 (\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2})$
or $( 1+i )^6 =-8i$

nth root of Complex number

If $Z=r(\cos \theta +i\sin \theta )$ be a complex number then the nth root of z is
$\sqrt[n]{Z}=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right )$
Proof
Given that Z is a complex number. Also let, nth root of Z is W such that $W=R(\cos \phi +i\sin \phi )$
Now we have
$\sqrt[n]{Z}=W$
or $W^n=Z$
or $[R(\cos \phi +i\sin \phi )]^n=r(\cos \theta +i\sin \theta )$
or $R^n(\cos (n\phi) +i\sin (n\phi) )=r(\cos \theta +i\sin \theta )$
Equating real and Imaginary parts, we get
$R^n=r$ and $\cos (n\phi)= \cos \theta$ and $\sin (n\phi) =\sin \theta$
or $R=\sqrt[n]{r}$ and $n\phi= \theta +2k \pi$
or $R=\sqrt[n]{r}$ and $\phi= \frac{(\theta +2k\pi )}{n}$
Thus, nth root of $Z=r(\cos \theta +i\sin \theta )$ is
$W=R(\cos \phi +i\sin \phi )$
or $W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right )$

square roots of i

Find the square roots of i

We know that
$\sqrt[n]{z}=\sqrt[n]{r} \left [ \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ]$

Solution
Since $i=\cos \frac{\pi }{2} +i\sin \frac{\pi }{2}$ , we get
r=1 and $\theta=\frac{\pi }{2}$
Hence the first square roots of i is
$w_1[i=0] =\cos \frac{\frac{\pi }{2}+0}{2} +i\sin \frac{\frac{\pi }{2}+0}{2} =\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$
The second square root of i is

$w_2[i=1] =\cos \frac{\frac{\pi }{2}+2\pi }{2} +i\sin \frac{\frac{\pi }{2}+2\pi }{2} =\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}=\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}$

Introduction

Meaning of i

Positive powers of i

Absolute Value

Conjugate

Properties of Conjugate numbers

Theorem: An important property

Question

Theorem: Triangle inequality:

Corollary

Algebra of complex number

Algebric Structure

Polar form

Example

Theorem

Some important property

Question

Exponenrial Form

Proof of the Formula

De-Moivre’s theorem

Example 1

nth root of Complex number

square roots of i

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Theorem: Triangle inequality: $| z_1 +z_2 |\le | z_1 |+| z_2 |$