Complex Numbers

Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. For example, 3+2i, -2+i√3 are complex numbers. For a complex number z = x+iy, x is called the real part, denoted by Re z and y is called the imaginary part denoted by Im z. For example, if z = 3+2i, Re z = 3 and Im z = 2.

In this section, aspirants will learn about complex numbers – definition, standard form, algebraic operations, conjugate, complex numbers polar form, Euler’s form and many more. A Complex Number is a combination of Real Number and an Imaginary Number. 

Table of Contents for Complex Numbers:

• Complex Numbers Definition
• Algebraic Operations
• Conjugate of Complex Number
• Properties
• Modulus and Argument
• Euler’s form
• Solved Problems

What are Complex Numbers?

If x, y ∈ R, then an ordered pair (x, y) = x + iy is called a complex number. It is denoted by z. Where x is real part of Re(z) and y is imaginary part or Im (z) of the complex number.

(i) If Re(z) = x = 0, then is called purely imaginary number

(ii) If Im(z) = y = 0 then z is called purely real number.

Note: The set of all possible ordered pairs is called complex number set, is denoted by C.

Integral powers of Iota

An imaginary number I (iota) is defined as √-1 since I = x√-1 we have i² = –1 , 1³ = –1, i⁴ = 1

• To find the value of in (n > 4) first, divide n by 4.

Let q is the quotient and r is the remainder.

n = 4q + r where o < r < 3

iⁿ = i^{4q + r} = (i4)^q , i^r = (1)^q . i^r = i^r

• The sum of four consecutive powers of I is zero.

In + in+1 + in+2 + in+3 = 0, n ∈ z

• 1/i = – i
• (1 + i)² = 2i and (1 – i)² = 2i
• √a . √b = √ab is valid only when atleast one of a and b is non negative.
• If both a and b both are negative then √a × √b = -√(|a|.|b|)
• √-a × √-b = -a

Watch this Video for More Reference

Illustration 1: Evaluate i²⁰¹

Solution: 201 leaves remainder as 1 when it is divided by 4 therefore i²⁰¹ = i¹ = i

Illustration 2: Evaluate 1 + (1+i) + (1+i)² + (1+i)³

Solution: 1 + (1+i) + (1+i)² + (1+i)³ = 1 + (1+i) + (2i) + (1+i)2i

= 1 + 1 + i + 2i + 2i³ = 5i

Illustration 3: [(1 + i)/√2]⁸ⁿ + [(1 – i)/√2]⁸ⁿ = ____

Solution:

= (2i/2)⁴ⁿ + (-2i/2)⁴ⁿ = i⁴ⁿ + (-i)⁴ⁿ

= 1 + 1 = 2

Illustration 4: Evaluate: (i⁴ⁿ⁺¹ – i^4n-1)/2, n ε z

Solution:

(i⁴ⁿ⁺¹ – i^{4n – 1})/2 = (i⁴ⁿ . i – i⁴ⁿ . i^-1)/2 = (i – i^-1)/2 = (i + i)/2 = i.

Algebraic Operations with Complex Numbers

1. Addition: (a + ib) + (c + id) = (a + c) + i(b + d)

2. Subtraction: (a + ib) – (c + id) = (ac) + i(b – d)

3. Multiplication: (a + ib) (c + id)

= (ac – bd) + i(ad + bc)

4. Reciprocal: If at least one of a, b is non-zero then the reciprocal of a + bi is given by

1/(a+ib) = (a – ib)/[(a+ib) (a−ib)] = a/[a² + b²] – i[b/(a² + b²)]

5. Quotient: If at least one of c, d is non-zero, then quotient of a + bi and c + di is given by

= [ac + bd]/[c² + d²] + i[bc−ad]/[c2+d2 ]

Conjugate of Complex Number

Let = z = a + ib be a complex number. We define conjugate of z, denoted by z¯ to be the complex number a – ib. that is, if z = a + ib, then z¯ = a – ib.

Properties of Conjugate of complex Number

(i) ${{z}_{1}}={{z}_{2}}\Leftrightarrow {{\overline{z}}_{1}}={{\overline{z}}_{2}}$

(ii) $\overline{(\bar{z})}=z$

(iii) $z+\overline{z}=2\,{Re}(z)$

(iv) $z-\overline{z}=2i\,{Im}\,(z)$

(v) $z=\overline{z}\Leftrightarrow z$ is purely real

(vi) $z+\overline{z}=0\Leftrightarrow z$ is purely imaginary.

(vii) $z\overline{z}={{[Re\,(z)]}^{2}}+{{[Im(z)]}^{2}}$

(viii) $\overline{{{z}_{1}}+{{z}_{2}}}={{\overline{z}}_{1}}+{{\overline{z}}_{2}}$

(ix) $\overline{{{z}_{1}}-{{z}_{2}}}={{\overline{z}}_{1}}-{{\overline{z}}_{2}}$

(x) $\overline{{{z}_{1}}{{z}_{2}}}={{\overline{z}}_{_{1}}}{{\overline{z}}_{_{2}}}$

(xi) $\overline{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}=\frac{{{\overline{z}}_{1}}}{{{z}_{2}}}$ if z₂ ≠ 0

(xii) If P(z) = a₀ + a₁ z + a₂ z² + …. + a_n zⁿ

Where a₀, a₁, ….. a_n and z are complex number, then $\overline{P(z)}={{\overline{a}}_{0}}+{{\overline{a}}_{1}}(\overline{z})+{{\overline{a}}_{2}}{{(\overline{z})}^{2}}+….+{{\overline{a}}_{n}}{{(\overline{z})}^{n}}$

= $\overline{P}(\overline{z})$

Where $\overline{P}(z)={{\overline{a}}_{0}}+{{\overline{a}}_{1}}z+{{\overline{a}}_{2}}{{z}^{2}}+….+{{\overline{a}}_{n}}{{z}^{n}}$

(xiii) If R(z) = $\frac{P(z)}{Q(z)}$ where P (z) and Q (z) are polynomials in z, and Q(z) ≠ 0, then

Modulus of a Complex Number

Let z = a + ib be a complex number. We define the modulus or the absolute value of z to be the real number √(a² + b²) and denote it by |z|.

Note that |z| > 0 ∀ z ∈ C

Properties of Modulus

If z is a complex number, then

(i) |z| = 0 ⇔ z = 0

(ii) |z| = |z¯| = |-z| = |-z¯|

(iii) – |z| ≤ Re (z) ≤ |z|

(iv) – |z| ≤ Im(z) ≤ |z|

(v) z z¯ = |z|²

If z₁, z₂ are two complex numbers, then

(vi) |z₁ z₂| = |z₁|.|z₂|

(vii) ∣z₁/z₂∣ = ∣z₁/z₂∣, if z₂ ≠ 0

(viii) |z₁ + z₂|² = |z₁|² + |z₂|² + z¯₁ z₂ + z₁ z^–₂  = |z₁|² + |z₂|² + 2Re (z₁ z¯₂)

(ix) |z₁+z₂|²  + |z₁|² – |z₂|² – z¯­₁ z₂ – z₁ z¯₂ = |z₁|² + |z₂|² – 2Re (z₁ z¯₂)

(x) |z₁+z₂|² + |z₁ – z₂|² = 2(|z₁|² + |z₂|²)

(xi) If a and b are real numbers and z₁, z₂ are complex numbers, then |az₁ + bz₂ |² + |bz₁ – az₂ |² = (a² + b²) (|z1|² + |z₂|²)

(xii) If z₁, z₂ ≠ 0, then |z₁ + z₂|² = |z₁|² + |z₂|² ⇔z₁ z₂ is purely imaginary.

(xiii) Triangle Inequality. If z₁ and z₂ are two complex numbers, then |z₁ + z₂| < |z₁| + |z₂|. The equality holds if and only if z₁ z¯2 ≥ 0.

In general, |z₁ + z₂+…+z_n| < |z₁| + |z₂| +…..+ |z_n| and the sign equality sign holds if and only if the ratio of any two non-zero terms is positive.

(xiv) |z₁ – z₂| ≤ |z₁| + |z₂|

(xv) ||z₁| – |z₂|| ≤ |z₁| + |z₂|

(xvi) |z₁ – z₂| ≥ ||z₁| – |z₂||

(xvii) If a₁, a₂, a₃, are four complex numbers, then |z – a₁| + |z – a₂| + |z – a₃| + |z – a₄| > max

{|a1−al|+|am−an| : l, m, n are distinct integers lying in{2, 3, 4}and m<n}.

Square Root of a Complex Number

Let z = x + iy then

Where |x| = $\sqrt{{{x}^{2}}+{{y}^{2}}}\,$

NOTE:

(i) $\sqrt{x+iy}+\sqrt{x-iy}=\sqrt{2|z|+2x}$

(ii) $\sqrt{x+iy}-\sqrt{x-iy}=i\sqrt{2|z|-2x}$

(iii) $\sqrt{i}=\pm \left( \frac{1+i}{\sqrt{2}} \right)\,and\,\sqrt{-i}=\pm \left( \frac{1-i}{\sqrt{2}} \right)$

Modulus and Argument of a Complex Number

Let z = x + iy = (x, y) for all x, y$\in$R and i = $\sqrt{-1}$

The length OP is called modulus of the complex number z denoted by |z|,

i.e. OP = r = |z| = $\sqrt{({{x}^{2}}+{{y}^{2}})}$

and if (x, y) ≠ (0, 0), then θ is called the argument or amplitude of z,

i.e. θ = ${{\tan }^{-1}}\left( \frac{y}{x} \right)$ [angle made by OP with positive X-axis]

or arg (z) = ${{\tan }^{-1}}\left( y/x \right)$

Also, argument of a complex number is not unique, since if θ is a value of the argument, so also in 2nπ + θ, where n $\in$ I. But usually, we take only that value for which

0 < θ < 2π. Any two arguments of a complex number differ by 2nπ.

Argument of z will be θ, π – θ, π + θ and 2π – θ according as the point z lies in I, II, III and IV quadrants respectively, where θ = ${{\tan }^{-1}}\left| \frac{y}{x} \right|$.

Illustration. Find the arguments of z₁ = 5 + 5i, z₂ = –4 + 4i, z₃ = –3 – 3i and z₄ = 2 – 2i, where

Solution: Since, z₁, z₂, z₃ and z₄ lies in I, II, III and IV quadrants respectively. The arguments are given by

arg (z₁) = ${{\tan }^{-1}}\left( \frac{5}{5} \right)={{\tan }^{-1}}1=\pi /4$

arg (z₂) = $\pi -{{\tan }^{-1}}\left| \frac{4}{-4} \right|=\pi -{{\tan }^{-1}}1=\pi -\frac{\pi}{4}=\frac{3\pi }{4}$

arg (z₃) = $\pi -{{\tan }^{-1}}\left| \frac{-3}{-3} \right|=\pi +{{\tan }^{-1}}1=\pi +\frac{\pi}{4}=\frac{5\pi }{4}$

And arg (z₄ ) = $2\pi -{{\tan }^{-1}}\left| \frac{-2}{2} \right|=2\pi +{{\tan }^{-1}}1=2\pi -\frac{\pi}{4}=\frac{7\pi }{4}$

Principal value of the Argument

The value θ of the argument which satisfies the inequality $-\pi <\theta \le \pi$ is called the principal value of the argument.

If x = x + iy = ( x, y), $\forall$ x, y $\in$ R and $i=\sqrt{-1}$, then

Arg(z) = ${{\tan }^{-1}}\left( \frac{y}{x} \right)$ always gives the principal value. It depends on the quadrant in which the point (x, y) lies.

(i) (x, y) $\in$ first quadrant x > 0, y > 0.

The principal value of arg (z) = $\theta ={{\tan }^{-1}}\left( \frac{y}{x} \right)$

It is an acute angle and positive.

(ii) (x, y) $\in$ second quadrant x < 0, y > 0.

The principal value of arg (z) = θ = $\pi -{{\tan }^{-1}}\left( \frac{y}{|x|}\right)$. It is an obtuse angle and positive.

(iii) (x, y) $\in$ third quadrant x < 0, y < 0.

The principal value of arg (z) = θ = $-\pi +{{\tan }^{-1}}\left( \frac{y}{x}\right)$

It is an obtuse angle and negative.

(iv) (x, y) $\in$ fourth quadrant x > 0, y < 0.

The principal value of arg (z) = θ = $-{{\tan }^{-1}}\left( \frac{|y|}{x}\right)$

It is an acute angle and negative.

Polar Form of a Complex Number

We have, z = x + iy

= |z| [cosƟ + i sinƟ]

Where |z| is the modulus of the complex number, ie., the distance of z from origin, and Ɵ is the argument or amplitude of the complex number.

Here we should take the principal value of Ɵ. For general values of argument z = r[cos(2nπ + Ɵ)] (where n is an integer). This is a polar form of the complex number.

Euler’s form of a Complex Number

e^iƟ = cos Ɵ + i sin Ɵ

This form makes the study of complex numbers and its properties simple. Any complex number can be expressed as

z = x + iy (Cartesian form)

= |z| [cos Ɵ + I sin Ɵ] (polar form)

= |z| e^iƟ

De Moivre’s Theorem and its Applications

(a) De Moivre’s Theorem for integral index. If n is a integer, then (cos Ɵ + i sin Ɵ)ⁿ = cos (nƟ) + I sin (nƟ)

(b) De Moivre’s Theorem for rational index. If n is a rational number, then value of or one of the values of

(cosƟ + isinƟ)ⁿ is cos (nƟ) + i sin (nƟ). In fact, if n = p/q where p, q ϵ I, q > 0 and p,q have no factors in

common, then (cos Ɵ + I sin Ɵ)n has q distinct values, one of which is cos (nƟ) + i sin (nƟ)

The nth Roots of Unity

By an nth root of unity we mean any complex number z which satisfies the equation zⁿ = 1 (1)

Since, an equation of degree n has n roots, there are n values of z which satisfy the equation (1). To obtain these n values of z, we write 1 = cos (2kπ) + I sin (2kπ)

Where k ϵ I and

Where k = 0, 1, 2, …., n -1.

Notation $\omega =\cos \left( \frac{2\pi }{n} \right)+i\sin \left( \frac{2\pi }{n} \right)$

By using the De Moivre’s theorem, we can write the nth roots of unity as 1, ω, ω², …., ω^n-1.

Sum of the Roots of Unity is Zero

We have 1 + ω + … + ω^{n – 1} = $\frac{1-{{\omega }^{n}}}{1-\omega }$

But ωⁿ = 1 as ω is a nth root of unity.

∴ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+\omega +…+{{\omega }^{n-1}}=0$

Also, note that

Cube roots of Unity

Cube roots of unity are given by 1, ω, ω², where $\omega =\cos \left( \frac{2\pi }{3} \right)+i\sin \left( \frac{2\pi }{3} \right)=\frac{-1+\sqrt{3i}}{2}and\,{{\omega }^{2}}=\frac{-1-\sqrt{3i}}{2}$

Some Results Involving Complex Cube Root of Unity (ω)

(i) ω³ = 1

(ii) 1 + ω + ω² = 0

(iii) x³ – 1 = (x – 1) ( x – ω) (x – ω²)

(iv) ω and ω² are roots of x² + x + 1 = 0

(v) a³ – b³ = (a – b) (a – bω) (a – bω²)

(vi) a² + b² + c² – bc – ca – ab

= (a + bω + cω²) (a + bω² + cω)

(vii) a³ + b³ + c³ – 3abc

= (a + b + c) (a + bω + cω²) (a + bω² + cω)

(viii) x³ + 1 = (x + 1) (x + ω) (x + ω²)

(ix) a³ + b³ = (a + b) (a + bω) (a + bω²)

(x) Cube roots of real number a are a^1/3, a^1/3ω, a^1/3 ω².

To obtain cube roots of a, we write x³ = a as y³ = 1 where y = x/a^1/3.

Solution of y³ = 1 are 1, ω, ω².

x = a^1/3, a^1/3 ω, a^1/3 ω².

Logarithm of a Complex Number

Log_e(x + iy) = log_e (|z|e^iƟ)

= loge |z| + log_e e^iƟ

= loge |z| + iƟ

= ${{\log }_{e}}\sqrt{({{x}^{2}}+{{y}^{2}})}+i\arg (z)$

∴ $\,\,\,\,\,\,\,\,\,{{\log }_{e}}(z)=lo{{g}_{e}}|z|+iarg(z)$

Problems on Complex Numbers

Illustration: The number of solutions of ${{z}^{3}}+\overline{z}=0$ is

(a) 2 (b) 3 (c) 4 (d) 5

Ans. (d)

Solution ${{z}^{3}}+\overline{z}=0\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{{z}^{3}}=-\overline{z}$ $\Rightarrow \,\,\,\,\,\,\,\,\,\,\,|z{{|}^{3}}=|-\overline{z}|\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,|z{{|}^{3}}=|z|$ $|z|\,(|z|-1)\,(|z|+1)=0$

⇒ |z|=0 or |z|=1 [Since, |z|+1>0]

If |z| = 0, then z = 0

If $|z|=1\,\,\,we\,get\,|z{{|}^{2}}=1\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,z\,\overline{z}=\,1$

Thus, ${{z}^{3}}+\overline{z}=0\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{3}}+1/z=0\,$ $\Rightarrow \,\,{{z}^{4}}+1=0$

This equation has four non-zero and distinct roots. Therefore, the given equation has five roots.

TIP It is unnecessary to find roots of z⁴ + 1 = 0

Illustration: If ω is an imaginary cube root of unity, then value of the expression

2(1 + ω) (1 + ω²) + 3(2 + ω) (2 + ω²) + … + (n + 1) (n + ω) (n + ω²) is

(a) $\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}+n$

(b) $\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}-n$

(c) $\frac{1}{4}n{{(n+1)}^{2}}-n$

(d) none of these

Ans. (a)

Solution r^th term of the given expression is

(r + 1) (r + ω) (r + ω²) = r³ + 1

Value of the given expression is

Illustration: Find the real part of ${{(1-i)}^{-i}}$

Sol: Let z = ${{(1-i)}^{-i}}$. Taking log on both sides, we have $log\,z=-i\,lo{{g}_{e}}(1-i)$

= $-i{{\log }_{e}}\sqrt{2}\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right)$

= $-i{{\log }_{e}}(\sqrt{2}{{e}^{-i}}^{(\pi /4)})$

= $-i\left[ \frac{1}{2}{{\log }_{e}}2+{{\log }_{e}}^{-i\pi /4} \right]$

= $-i\left[ \frac{1}{2}{{\log }_{e}}2-\frac{i\pi }{4} \right]$

= $-\frac{i}{2}{{\log }_{e}}2-\frac{\pi }{4}$ $\Rightarrow \,\,\,\,\,\,\,\,\,\,\,z={{e}^{-\pi /4}}\,{{e}^{-i(log\,2)/2}}$ $\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{Re}(z)={{e}^{-\pi /4}}\cos \left( \frac{1}{2}\log 2 \right)$

Illustration: If α, β and γ are the roots of x³ – 3x² + 3x + 7 = 0, find the value of

Sol. We have, x³ – 3×2 + 3x + 7 = 0

∴ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-1,1-2\omega ,1-2{{\omega }^{2}}$ $\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha =-1,\beta =1-2\omega ,\gamma=1-2{{\omega}^{2}}$

Then, $\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}=\frac{-2}{-2\omega }+\frac{-2\omega }{-2{{\omega }^{2}}}+\frac{-2{{\omega }^{2}}}{-2}$ $=\frac{1}{\omega }+\frac{1}{\omega }+{{\omega }^{2}}={{\omega }^{2}}+{{\omega }^{2}}+{{\omega }^{2}}=3{{\omega }^{2}}$

Illustration: $f\,\,{{z}_{1}}\,\,and\,{{z}_{2}}\,are\,\,1-i,-2+4i$ respectively. Find ${Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{z}_{1}}} \right).$

Sol. $\frac{{{z}_{1}}{{z}_{2}}}{{{z}_{1}}}=\frac{\left( 1-i \right)\left( -2+4i \right)}{1+i}=\frac{-2+2i+4i+4}{1+i}$ $=\frac{2+6i}{1+i}\times \frac{1-i}{1-i}=\frac{2+6i-2i+6}{2}=4+2i$

∴ $\,\,\,\,\,{Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}} \right)=2.$

Illustration: Find the square root of $z=-7-24i.$

Sol. Consider ${{z}_{0}}=x+iy$ be a square root then ${{z}_{0}}^{2}=-7-24i.$ $-7-24i={{x}^{2}}-{{y}^{2}}+2ixy$

Equating real and imaginary parts we get

and $2xy=-24$ ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}}$ $={{\left( -7 \right)}^{2}}+{{\left( -24 \right)}^{2}}=625$

∴ ${{x}^{2}}+{{y}^{2}}=25$

Solving (i) and (iii), we get,

∴ ${{z}_{0}}=\pm \left( 3-4i \right)$

Illustration: If n is a positive integer and $\omega$ be an imaginary cube root of unity, prove that

Sol. Case: I. $n=3m;m\in I$

∴$1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{3m}}+{{\omega }^{6m}}$ $=1+1+1\left[ Since,\;\;{{\omega }^{3}}=1 \right]=3$

Case: II. $n=3m+1\,or\,3m+2;m\in I$

(a) Let $n=3m+1$

∴ $L.H.S=1{{\omega }^{3m+1}}+{{\omega }^{6m+2}}=1+\omega +{{\omega }^{2}}=0$

(b) Let $n=3m+2$ $1+{{\omega }^{3m+2}}+{{\omega }^{6m+4}}=1+{{\omega }^{2}}+{{\omega }^{4}}=1+{{\omega }^{2}}+\omega =0.$

Illustration: Show that $\left| \frac{z-3}{z+3} \right|=2$ represents a circle.

Sol. Consider $z=x+iy$

∴$\left| \frac{z-3}{z+3} \right|=2\Rightarrow \left| \frac{x-3+iy}{x+3+iy} \right|=2$ ${{\left| x-3+iy \right|}^{2}}={{2}^{2}}{{\left| x+3+iy \right|}^{2}}$

or ${{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\left( {{\left( x+3 \right)}^{2}}+{{y}^{2}} \right)$ $\Rightarrow 3{{x}^{2}}+3{{y}^{2}}+30x+27=0$

which represents a circle.

Illustration: If $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=…….=\left| {{z}_{n}} \right|=1$

Prove that $\left| {{z}_{1}}{{z}_{2}}+…….+{{z}_{n}} \right|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+…….+\frac{1}{{{z}_{n}}} \right|$

Sol. $\left| {{z}_{j}} \right|=1\Rightarrow {{z}_{j}}{{\bar{z}}_{j}}=1\forall j=1,……,n$ $\left( Since, \;z\bar{z}=\left| {{z}^{2}} \right| \right)$

L.H.S.

Illustration: If $\left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|,$ prove that $\arg {{z}_{1}}-\arg {{z}_{2}}=$ odd multiple of $\frac{\pi }{2}.$

Sol. As we know $\left| z \right|=z.\bar{z}.$ Apply this formula and consider $z=r\left( \cos \theta +i\,sin\theta \right).$ ${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}$ $\Rightarrow \left( {{z}_{1}}+{{z}_{2}} \right)\left( {{{\bar{z}}}_{1}}+{{{\bar{z}}}_{2}} \right)=\left( {{z}_{1}}-{{z}_{2}} \right)\left( {{{\bar{z}}}_{1}}-{{{\bar{z}}}_{2}} \right)\,\,or$ ${{z}_{1}}{{\bar{z}}_{1}}+{{z}_{2}}{{\bar{z}}_{2}}+{{z}_{2}}{{\bar{z}}_{1}}+{{z}_{1}}{{\bar{z}}_{2}}={{z}_{1}}{{\bar{z}}_{1}}+{{z}_{2}}{{\bar{z}}_{2}}-{{z}_{2}}{{\bar{z}}_{1}}-{{z}_{1}}{{\bar{z}}_{2}}$

or $2\left( {{z}_{2}}{{{\bar{z}}}_{1}}+{{z}_{1}}{{{\bar{z}}}_{2}} \right)=0;{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)=0$

Let ${{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\,\sin {{\theta }_{1}} \right)\,and\,{{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right);$

then ${{z}_{1}}{{\bar{z}}_{2}}={{r}_{1}}{{r}_{2}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right)$

∴ $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=0\left( as\,{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)=0 \right)$ ${{\theta }_{1}}-{{\theta }_{2}}=$ odd multiple of $\frac{\pi }{2}.$

Illustration: If |z – 1| < 3, prove that |iz + 3 – 5i| < 8.

Sol: Here we have to reduce iz + 3 – 5i as the sum of two complex numbers containing z – 1, because we have to use

|z – 1| < 3.

|iz + 3 – 5i| = |iz – i + 3 – 4i| = |3 – 4i + i (z – 1) | < |3 – 4i| + |i (z – 1 )|

(by triangle inequality) < 5 + 1 . 3 = 8

Illustration: If (1 + x)ⁿ = a₀ + a₁x + a₂x²+ ….+ a_nxⁿ, then show that

(a) ${{a}_{0}}-{{a}_{2}}+{{a}_{4}}+….={{2}^{\frac{n}{2}}}\cos \frac{n\pi }{4}$

(b) ${{a}_{1}}-{{a}_{3}}+{{a}_{5}}+….={{2}^{\frac{n}{2}}}\sin \frac{n\pi }{4}$

Sol: simply put x = i in the given expansion and then by using formula

z = r (cosƟ + i sinƟ) and (cosƟ + i sin Ɵ)ⁿ

= cosnƟ + i sin nƟ, we can solve this problem.

Put x = I in the given expansion

(1 + i)ⁿ = a₀ + a₁i + a₂i² + ….+ a_niⁿ.

Equating real and imaginary parts.

Example 9: Solve the equation ${{z}^{n-1}}=\overline{z}:n\in N$

Sol: Apply modulus on both side.

∴ $\,\,|z|=0\,or\,|z|\,=1\,\,If\,|z|=0\,then\,z\,=\,0,$ $Let\left| z \right|=1;\text{ }then,\text{ }{{z}^{n}}=z\overline{z}=1$

∴ $\,\,\,\,\,\,\,\,z=\cos \frac{2m\pi }{n}+\sin \frac{2m\pi }{n}:m=0,\,1,\,…..,\,n-1$

Illustration: If z = x + iy and $\omega =\frac{1-iz}{z-i}$

with |ω| = 1, show that, z ; lies on the real axis.

Sol: Substitute value of ω in |ω| = 1

or, (1 + y)² + x² = x² + (y – 1)² or, 4y = 0

Hence z lies on the real axis.

Illustration: If a complex number z lies in the interior or on the boundary of a circle of radius as 3 and centre at (0, –4) then greatest and least value of |z + 1| are-

(a) $3+\sqrt{17},\sqrt{17}-3$

(b) 6, 1

(c) $\sqrt{17},1$

(d) 3, 1

Sol: Greatest and least value of |z + 1| means maximum and minimum distance of circle from the point (–1, 0). In a circle, greatest and least distance of it from any point is along the normal.

∴ $\,\,\,\,Greatest\,dis\tan ce\,=\,3+\sqrt{{{1}^{2}}+{{4}^{2}}}=3+\sqrt{17}$

Least distance = $\sqrt{{{1}^{2}}+{{4}^{2}}}-3=\sqrt{17}-3$

Illustration: Find the equation of the circle for which arg $\left( \frac{z-6-2i}{z-2-2i} \right)=\pi /4.$

Sol: $\arg \left( \frac{z-6-2i}{z-2-2i} \right)=\pi /4$

represent a major arc of circle of which Line joining (6, 2)) and (2, 2) is a chord that subtends an angle $\frac{\pi }{4}$ at circumference.

Clearly AB is parallel to real (x) axis, M is mid-point, M ≡ (4, 2), OM = AM = 2

∴ $\,\,O=(4,4)\,and\,O{{A}^{2}}=O{{M}^{2}}+A{{M}^{2}}=2\sqrt{2}$ Equation of required circle is

Illustration: if |z| > 3, prove that the least value of $\left| z+\frac{1}{z} \right|\,is\,\frac{8}{3}$

Sol: Here $\left| z+\frac{1}{z} \right|\,\underline{>}\,|z|-\frac{1}{|z|}$ Now |z| > 3

∴ $\,\,\,\,\frac{1}{|z|}\underline{<}\frac{1}{3}\,or\,-\frac{1}{|z|}\underline{>}-\frac{1}{3}$ (i)

Adding the two like inequalities