Representation of a Complex Number

A complex number is a number that can be expressed in the form of a + ib where a represents the real part and b is the imaginary part; i is the imaginary unit which is defined as the square root of -1 or we can have i as the solution of x2 = -1. In this article, you will learn the representation of a complex number.

A complex number whose real part is zero is said to be a purely imaginary number and the points of these numbers lie on the vertical axis of the complex plane. Similarly, a complex number whose imaginary part is zero can be viewed as a real number and its point lies on the horizontal axis of the complex plane.

Complex Number Geometrical Representation

Geometrically, complex numbers extend the concept of one-dimensional number line to the two-dimensional complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. Thus, if given a complex number a+bi, it can be identified as a point P(a,b) in the complex plane. Complex numbers can also be represented in Polar form, that associates each complex number with its distance from the origin as its magnitude and with a particular angle and this is called as the argument of the complex number.

Plotting a Complex Number

To plot a complex number a + ib which can be thought of as a point P(a,b), we consider s-plane similar to x-y plane relating to the normal cartesian system.

S-plane is also called as Argand plane or Complex-plane or Argand diagram, named after Jean-Robert Argand plots the real and imaginary values a and b respectively of the point P on the horizontal axis called as the real axis and on the vertical axis called as Imaginary axis.

Consider a point (2,3) in complex plane. It can be defined as 2+3i in terms of a complex number, plotted in the first quadrant as both the real and imaginary terms are positive. Similarly, we can have -2+3i as (-2,3) in second quadrant; (-2, -3) as -2-3i in third quadrant and (2, -3) as 2-3i in fourth quadrant.

Polar Complex Plane

Complex Number Representation

Above figure represents a line joining some points in the complex plane, whose length is r and makes an angle Ɵ with the real axis. From the line, draw a line projecting parallel to Real axis and meeting the Imaginary axis and let it be b. Similarly, a line drawn from the plot touching Real axis parallel to imaginary axis be a. Then, by Pythagoras theorem, we have r = a2+b2\sqrt{a^2 +b^2} which is often referred to as magnitude. It is also referred to as the absolute value or modulus of the complex number, or r = |z|. Similarly, we will have the argument, tan Ɵ = b/a or simply Ɵ = tan-1(y/x). It is often referred to as Phase or angle of the complex number.

If the complex number has no imaginary part in it or b=0; then r=|x|, i.e., the absolute value of the real number equals the absolute value of the complex number.

Consider the polar form z = r cos Ɵ + r i sin Ɵ

=> z = r (cos Ɵ + i sin Ɵ)

By using Euler’s formula, we have z = r e

=> z= r < Ɵ

Also Read

Modulus And Conjugate Of A Complex Number

Important Complex Numbers Properties

  • For two complex numbers z1 and z2 to be equal, the corresponding real and imaginary parts of both the complex numbers are supposed to be equal.

Consider a complex number z1 = a + ib and z2 = c + id;

then if z1 = z2 => a = c and b=d.

In General, Re(z1) = Re(z2) and Im(z1) = Im(z2) implies z1 = z2.

  • Conjugate of a complex number

Consider a complex number z = a + ib. Then its conjugate is expressed as z* = a – ib. It is often called the reflection of z about the real axis.

Note: Conjugating a complex number twice results in the complex number itself, i.e., z** = z.

Consider z = a+ib. then, z* = a – ib and z** = a + ib; thus z** = z is justified.

Moreover, Re(z*) = Re(z) and |z*| = |z|.

Similarly, Im(z*) = -Im(z) and arg(z*) = -arg(z).

zz* = |z|2 = |z*|2

Re(z) = (z+z*)/2

Im(z) = (z-z*)/2i

  • Arithmetic Operation on Complex Numbers

Consider z1= a + ib and z2 = c + id;

Addition: z1 + z2 = (a + c) + i(b + d)

Subtraction: z1 – z2 = (a – c) + i(b – d)

Multiplication: z1 * z2 = ac b d + i(ad + bc).

Division: To work with complex division, we need to take the conjugate of the denominator and multiply both numerator and denominator by the conjugate value.

Let z = a+idc+id=ac+bd+i(bcad)c2+d2\frac{a+id}{c+id} = \frac{ac+bd + i(bc – ad)}{c^2 + d^2}

  • Complex Number Commutative Property

z1 + z2 = z2 + z1

z1 * z2 = z2 * z1

|z1 + z2 | ≤ |z1| + |z2 |

  • Construction as ordered pairs:

(a,b) + (c,d) = (a + c, b + d)

(a,b) . (c,d) = (ac b d, bc + ad)

Important Concepts related to Complex Numbers

Complex Analysis

The study of functions of a complex variable is known as Complex Analysis. They have four-dimensional graphs and may be usefully illustrated by color-coding a three-dimensional graph to suggest four dimensions.

Holomorphic Functions

A function f: C -> C is called holomorphic if it satisfies the Cauchy-Riemann equations.

Consider f(z) = u z + v z*, with complex coefficients u and v. This map is holomorphic if and only if b = 0.

Solved Problems

Problem 1: Represent the complex number z = 1+ i√3 in the polar form.

Solution:

Let r cos θ = 1 and r sin θ = √3.

Squaring and adding, we get

r2 (cos2  θ +sin2  θ ) = 1+3

r2 = 4

i.e. r = 2

Substitute r in r cos θ = 1

Therefore cos θ = 1/2

Also sin θ = √3/2

⇒ θ = π/3

Hence the required polar form is z = 2 (cos  π/3 + i sin π/3).

Problem 2: Let a complex number be defined by z = 2 + 5i and another complex number a = 3 + xi and b = y + 2i. Given that z = a + b. Then, find the values of x and y?

Solution: z = a + b -> (2 + 5i) = (3 + xi) + (y + 2i)

=> 2 = 3 + y and 5 = x + 2

=> Solving, we have x = 3 and y = -1.

Problem 3: Find the conjugate of z1 if z2 – z3 = 0 and z1 = 3 * z3; given z2 = 5 + 5i and z3 = x + yi? Also, find the magnitude and argument of conjugate of z2?

Solution: z2 – z3 = 0 implies 5 + x = 0 and 5 + y = 0.

Solving we have x – 5 and y = -5.

Thus, z3 = -5 – 5i.

Given z1 = 3 * z3

=> 3(-5 – 5i) = -15 -15i.

We need to find the complex conjugate of z1.

Hence, z1* = -15 + 15i.

Now, z2 * = 5 – 5i.

|z2 |= (25 + 25)0.5 = 7.07

And Ɵ = tan-1 (y/x) = tan-1 (-5/5) = tan-1 (-1) = -450.

Problem 4:  ilog(xix+i)i\log \left( \frac{x-i}{x+i} \right) is equal to

Solution:

Let 

z=ilog(xix+i)zi=log(xix+i)zi=log[xix+i×xixi]=log[x212ixx2+1]zi=log[x21x2+1i2xx2+1](i) because log(a+ib)=log(reiθ)=logr+iθ=loga2+b2+itan1(b/a)z=\,i\log \left( \frac{x-i}{x+i} \right) \\ \Rightarrow \,\frac{z}{i}=\log \left( \frac{x-i}{x+i} \right)\\ \Rightarrow \frac{z}{i}=\log \,\left[ \frac{x-i}{x+i}\times \frac{x-i}{x-i} \right]\\ =\,\log \,\left[ \frac{{{x}^{2}}-1-2ix}{{{x}^{2}}+1} \right]\\ \Rightarrow \frac{z}{i}=\log \left[ \frac{{{x}^{2}}-1}{{{x}^{2}}+1}-i\frac{2x}{{{x}^{2}}+1} \right] \rightarrow (i) \text \ because \ \,\log (a+ib)=\log (r{{e}^{i\theta }})=\log r+i\theta \\ = \log \sqrt{{{a}^{2}}+{{b}^{2}}}+i{{\tan }^{-1}}(b/a)

Hence,

zi=log(x21x2+1)2+(2xx2+1)2+itan1(2xx21)zi=logx4+12x2+4x2(x2+1)2+itan1(2x1x2)=log1+i(2tan1x)=0+i(2tan1x) therefore z=i22tan1x=2tan1x=π2tan1x.\frac{z}{i}=\log \sqrt{{{\left( \frac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)}^{2}}+{{\left( \frac{-2x}{{{x}^{2}}+1} \right)}^{2}}}+i{{\tan }^{-1}}\left( \frac{-2x}{{{x}^{2}}-1} \right) \\ \frac{z}{i}=\log \frac{\sqrt{{{x}^{4}}+1-2{{x}^{2}}+4{{x}^{2}}}}{{{({{x}^{2}}+1)}^{2}}}+i{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\\ =\log 1+i\,(2{{\tan }^{-1}}x)\\ =0+i\,(2{{\tan }^{-1}}x) \text \ therefore \ z={{i}^{2}}2{{\tan }^{-1}}x=-2{{\tan }^{-1}}x \\ =\pi -2{{\tan }^{-1}}x.

Problem 5:   (1+i3)20{{(-1+i\sqrt{3})}^{20}} is equal to

Solution:

Let 

z=1+i3,r=1+3=2θ=tan1(31)=2π3 therefore z=2(cos2π3+isin2π3) therefore (z)20=[2(cos2π3+isin2π3)]20=220(cos2π3+isin2π3)20=220(12+i32)20.z=-1+i\sqrt{3}, r=\sqrt{1+3}=2\\ \theta ={{\tan }^{-1}}\left( \frac{\sqrt{3}}{-1} \right)=\frac{2\pi }{3}\\ \text \ therefore \ \,z=2\,\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right) \\ \text \ therefore \ \,\,{{(z)}^{20}}={{\left[ 2\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right) \right]}^{20}}\\ ={{2}^{20}}{{\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)}^{20}}\\ ={{2}^{20}}{{\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)}^{20}}.

Problem 6: If z=1+i33+i,z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}, then (zˉ)100{{(\bar{z})}^{100}} lies in 

A) I quadrant

B) II quadrant

C) III quadrant

D) IV quadrant

Solution:

z=1+i33+iz=1+i33+i×3i3iz=3+3ii+33+1=2(3+i)4z=3+i2=[cosπ6+isinπ6] Now zˉ=cosπ6isinπ6(zˉ)100=[cosπ6isinπ6]100(zˉ)100=cos50π3isin50π3=cos2π3isin2π3(zˉ)100z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\\ \Rightarrow z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\times \frac{\sqrt{3}-i}{\sqrt{3}-i}\\ \Rightarrow z=\frac{\sqrt{3}+3i-i+\sqrt{3}}{3+1}\\ =\frac{2(\sqrt{3}+i)}{4}\\ \Rightarrow z=\frac{\sqrt{3}+i}{2}=\left[ \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right] \\ \text \ Now \ \bar{z}=\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\\ {{(\bar{z})}^{100}}={{\left[ \cos \frac{\pi }{6}-i\sin \frac{\pi }{6} \right]}^{100}}\\ {{(\bar{z})}^{100}}=\cos \frac{50\,\pi }{3}-i\sin \frac{50\,\pi }{3}\\ =\,\cos \frac{2\pi }{3}-i\sin \frac{2\pi }{3}\\ {{(\bar{z})}^{100}} 

lies in III quadrant.

Problem 7: The amplitude of eeiθ{{e}^{{{e}^{-i\theta }}}} is equal to

A)sinθB)sinθC)ecosθD)esinθA)\sin \theta \\ B)-\sin \theta \\ C){{e}^{\cos \theta }}\\ D){{e}^{\sin \theta }}

Solution:

Let

z=eeiθ=ecosθisinθ=ecosθeisinθz=ecosθ[cos(sinθ)isin(sinθ)]z=ecosθcos(sinθ)iecosθsin(sinθ)amp(z)=tan1[ecosθsin(sinθ)ecosθcos(sinθ)]=tan1[tan(sinθ)]=sinθ.z={{e}^{{{e}^{-i\theta }}}}={{e}^{\cos \theta -i\sin \theta }}\\ ={{e}^{\cos \theta }}{{e}^{-i\sin \theta }}\\ z={{e}^{\cos \theta }}[\cos (\sin \theta )-i\sin (\sin \theta )]\\ z={{e}^{\cos \theta }}\cos (\sin \theta )-i{{e}^{\cos \theta }}\sin (\sin \theta )\\ amp(z)={{\tan }^{-1}}\left[ -\frac{{{e}^{\cos \theta }}\sin (\sin \theta )}{{{e}^{\cos \theta }}\cos (\sin \theta )} \right]\\ ={{\tan }^{-1}}[\tan (-\sin \theta )]=-\sin \theta.