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:

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 = i4q + r = (i4)q , ir = (1)q . ir = ir

  • 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:

[(1 + i)/√2]8n + [(1 – i)/√2]8n = [{(1 + i)/√2}2]4n + [{(1 – i)/√2}2]4n

= (2i/2)4n + (-2i/2)4n = i4n + (-i)4n

= 1 + 1 = 2

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

Solution:

(i4n+1 – i4n – 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

[(a+bi)/(c+di)] = [(a+ib) (c−id)]/[(c+id) (c−id)] = [(ac + bd) + i(bc – ad)]/[c2 + d2]

= [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) z1=z2z1=z2{{z}_{1}}={{z}_{2}}\Leftrightarrow {{\overline{z}}_{1}}={{\overline{z}}_{2}}

(ii) (zˉ)=z\overline{(\bar{z})}=z

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

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

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

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

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

(viii) z1+z2=z1+z2\overline{{{z}_{1}}+{{z}_{2}}}={{\overline{z}}_{1}}+{{\overline{z}}_{2}}

(ix) z1z2=z1z2\overline{{{z}_{1}}-{{z}_{2}}}={{\overline{z}}_{1}}-{{\overline{z}}_{2}}

(x) z1z2=z1z2\overline{{{z}_{1}}{{z}_{2}}}={{\overline{z}}_{_{1}}}{{\overline{z}}_{_{2}}}

(xi) (z1z2)=z1z2\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 + …. + an zn

Where a0, a1, ….. an and z are complex number, then P(z)=a0+a1(z)+a2(z)2+.+an(z)n\overline{P(z)}={{\overline{a}}_{0}}+{{\overline{a}}_{1}}(\overline{z})+{{\overline{a}}_{2}}{{(\overline{z})}^{2}}+….+{{\overline{a}}_{n}}{{(\overline{z})}^{n}}

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

Where P(z)=a0+a1z+a2z2+.+anzn\overline{P}(z)={{\overline{a}}_{0}}+{{\overline{a}}_{1}}z+{{\overline{a}}_{2}}{{z}^{2}}+….+{{\overline{a}}_{n}}{{z}^{n}}

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

R(z)=P(z)Q(z)\overline{R\,(z)}=\frac{\overline{P}(\overline{z})}{\overline{Q}(\overline{z})}

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 z2  = |z1|2 + |z2|2 + 2Re (z12)

(ix) |z1+z2|2  + |z1|2 – |z2|2 – z¯­1 z2 – z12 = |z1|2 + |z2|2 – 2Re (z1 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, |z+ z2+…+zn| < |z1| + |z2| +…..+ |zn| 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

x+iy={±[z+x2+izx2]ify>0±[z+x2izx2]ify<0\sqrt{x+iy}=\left\{ \begin{matrix} \pm \left[ \sqrt{\frac{|z|+x}{2}}+i\sqrt{\frac{|z|-x}{2}} \right]if\,y\,>\,0 \\ \pm \left[ \sqrt{\frac{|z|+x}{2}}-i\sqrt{\frac{|z|-x}{2}} \right]if\,y\,<\,0 \\ \end{matrix} \right.

Where |x| = x2+y2\sqrt{{{x}^{2}}+{{y}^{2}}}\,

NOTE:

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

(ii) x+iyxiy=i2z2x\sqrt{x+iy}-\sqrt{x-iy}=i\sqrt{2|z|-2x}

(iii) i=±(1+i2)andi=±(1i2)\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\inR and i = 1\sqrt{-1}

 

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

i.e. OP = r = |z| = (x2+y2)\sqrt{({{x}^{2}}+{{y}^{2}})}

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

i.e. θ = tan1(yx){{\tan }^{-1}}\left( \frac{y}{x} \right) [angle made by OP with positive X-axis]

or arg (z) = tan1(y/x){{\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 = tan1yx{{\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

i=1i=\sqrt{-1}.

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

arg (z1) = tan1(55)=tan11=π/4{{\tan }^{-1}}\left( \frac{5}{5} \right)={{\tan }^{-1}}1=\pi /4

arg (z2) = πtan144=πtan11=ππ4=3π4\pi -{{\tan }^{-1}}\left| \frac{4}{-4} \right|=\pi -{{\tan }^{-1}}1=\pi -\frac{\pi}{4}=\frac{3\pi }{4}

arg (z3) = πtan133=π+tan11=π+π4=5π4\pi -{{\tan }^{-1}}\left| \frac{-3}{-3} \right|=\pi +{{\tan }^{-1}}1=\pi +\frac{\pi}{4}=\frac{5\pi }{4}

And arg (z4 ) = 2πtan122=2π+tan11=2ππ4=7π42\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 π<θπ-\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=1i=\sqrt{-1}, then

Arg(z) = tan1(yx){{\tan }^{-1}}\left( \frac{y}{x} \right) always gives the principal value. It depends on the quadrant in which the point (x, y) lies.

Principal value of argument

 

 

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

The principal value of arg (z) = θ=tan1(yx)\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 = πtan1(yx)\pi -{{\tan }^{-1}}\left( \frac{y}{|x|}\right). It is an obtuse angle and positive.Principal value of the argument second quadrant

 

 

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

The principal value of arg (z) = θ = π+tan1(yx)-\pi +{{\tan }^{-1}}\left( \frac{y}{x}\right)

Principal value of the argument third quadrant

 

It is an obtuse angle and negative.

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

The principal value of arg (z) = θ = tan1(yx)-{{\tan }^{-1}}\left( \frac{|y|}{x}\right)

Principal value of the argument fourth quadrant

 

It is an acute angle and negative.

Polar Form of a Complex Number

We have, z = x + iy

=x2+y2[2x2+y2+ixx2+y2]=\sqrt{{{x}^{2}}+{{y}^{2}}}\left[ \frac{2}{\sqrt{{{x}^{2}}+{{y}^{2}}}}+i\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right]

= |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 = 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

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)

Note

The values of (cos Ɵ + I sin Ɵ)p/q where p, q ϵ I, q ≠ 0, hcf (p,q) = 1 are given by

cos[pq(2kπ+θ)]+isin[pq(2kπ+θ)]\cos \left[ \frac{p}{q}(2k\pi +\theta ) \right]+i\,sin\left[ \frac{p}{q}(2k\pi +\theta ) \right]

Where k = 0, 1, 2, ….., q -1.

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

z=cos(2kπn)+isin(2kπn)\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,z=\cos \left( \frac{2k\pi }{n} \right)+i\sin \left( \frac{2k\pi }{n} \right) [using the De Moivre’s Theorem]

Where k = 0, 1, 2, …., n -1.

Note

We may give any n consecutive integral values to k. For instance, in case of 3, we may take -1, 0 and 1 and in case of 4, we may take – 1, 0, 1 and 2 or -2, -1, 0 and 1.

Notation ω=cos(2πn)+isin(2πn)\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, …., ωn-1.

Sum of the Roots of Unity is Zero

We have 1 + ω + … + ωn – 1 = 1ωn1ω\frac{1-{{\omega }^{n}}}{1-\omega }

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

1+ω++ωn1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+\omega +…+{{\omega }^{n-1}}=0

Also, note that

1x1+1xω.+1xωn1=nxn1xn1\frac{1}{x-1}+\frac{1}{x-\omega }….+\frac{1}{x-{{\omega }^{n-1}}}=\frac{n{{x}^{n-1}}}{{{x}^{n}}-1}

Cube roots of Unity

Cube roots of unity are given by 1, ω, ω2, where ω=cos(2π3)+isin(2π3)=1+3i2andω2=13i2\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 a1/3, a1/3ω, a1/3 ω2.

To obtain cube roots of a, we write x3 = a as y3 = 1 where y = x/a1/3.

Solution of y3 = 1 are 1, ω, ω2.

x = a1/3, a1/3 ω, a1/3 ω2.

Logarithm of a Complex Number

Loge(x + iy) = loge (|z|e)

= loge |z| + loge e

= loge |z| + iƟ

= loge(x2+y2)+iarg(z){{\log }_{e}}\sqrt{({{x}^{2}}+{{y}^{2}})}+i\arg (z)

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

Problems on Complex Numbers

Illustration: The number of solutions of z3+z=0{{z}^{3}}+\overline{z}=0 is

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

Ans. (d)

Solution z3+z=0z3=z{{z}^{3}}+\overline{z}=0\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{{z}^{3}}=-\overline{z} z3=zz3=z\Rightarrow \,\,\,\,\,\,\,\,\,\,\,|z{{|}^{3}}=|-\overline{z}|\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,|z{{|}^{3}}=|z| z(z1)(z+1)=0|z|\,(|z|-1)\,(|z|+1)=0

⇒ |z|=0 or |z|=1 [Since, |z|+1>0]

If |z| = 0, then z = 0

If z=1wegetz2=1zz=1|z|=1\,\,\,we\,get\,|z{{|}^{2}}=1\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,z\,\overline{z}=\,1

Thus, z3+z=0z3+1/z=0{{z}^{3}}+\overline{z}=0\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{3}}+1/z=0\, z4+1=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) 14n2(n+1)2+n\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}+n

(b) 14n2(n+1)2n\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}-n

(c) 14n(n+1)2n\frac{1}{4}n{{(n+1)}^{2}}-n

(d) none of these

Ans. (a)

Solution rth term of the given expression is

(r + 1) (r + ω) (r + ω2) = r3 + 1

Value of the given expression is

r=1n(r3+1)=14n2(n+1)2+n\sum\limits_{r=1}^{n}{({{r}^{3}}+1)}=\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}+n

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

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

= iloge2(cosπ4isinπ4)-i{{\log }_{e}}\sqrt{2}\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right)

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

= i[12loge2+logeiπ/4]-i\left[ \frac{1}{2}{{\log }_{e}}2+{{\log }_{e}}^{-i\pi /4} \right]

= i[12loge2iπ4]-i\left[ \frac{1}{2}{{\log }_{e}}2-\frac{i\pi }{4} \right]

= i2loge2π4-\frac{i}{2}{{\log }_{e}}2-\frac{\pi }{4} z=eπ/4ei(log2)/2\Rightarrow \,\,\,\,\,\,\,\,\,\,\,z={{e}^{-\pi /4}}\,{{e}^{-i(log\,2)/2}} Re(z)=eπ/4cos(12log2)\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

α1β1+β1γ1+γ1α1.\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}.

Sol. We have, x3 – 3×2 + 3x + 7 = 0

(x1)3+8=0\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(x-1)}^{3}}+8=0 (x1)3+23=0\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(x-1)}^{3}}+{{2}^{3}}=0 (x1+2)(x1+2ω)(x1+2ω2)=0\Rightarrow \,\,\,(x-1+2)(x-1+2\omega )(x-1+2{{\omega }^{2}})=0 (x+1)(x1+2ω)(x1+2ω2)=0\Rightarrow \,\,\,(x+1)(x-1+2\omega )(x-1+2{{\omega }^{2}})=0

x=1,12ω,12ω2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-1,1-2\omega ,1-2{{\omega }^{2}} α=1,β=12ω,γ=12ω2\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha =-1,\beta =1-2\omega ,\gamma=1-2{{\omega}^{2}}

Then, α1β1+β1γ1+γ1α1=22ω+2ω2ω2+2ω22\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} =1ω+1ω+ω2=ω2+ω2+ω2=3ω2=\frac{1}{\omega }+\frac{1}{\omega }+{{\omega }^{2}}={{\omega }^{2}}+{{\omega }^{2}}+{{\omega }^{2}}=3{{\omega }^{2}}

Illustration: fz1andz2are1i,2+4if\,\,{{z}_{1}}\,\,and\,{{z}_{2}}\,are\,\,1-i,-2+4i respectively. Find Im(z1z2z1).{Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{z}_{1}}} \right).

Sol. z1z2z1=(1i)(2+4i)1+i=2+2i+4i+41+i\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} =2+6i1+i×1i1i=2+6i2i+62=4+2i=\frac{2+6i}{1+i}\times \frac{1-i}{1-i}=\frac{2+6i-2i+6}{2}=4+2i

Im(z1z2zˉ1)=2.\,\,\,\,\,{Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}} \right)=2.

Illustration: Find the square root of z=724i.z=-7-24i.

Sol. Consider z0=x+iy{{z}_{0}}=x+iy be a square root then z02=724i.{{z}_{0}}^{2}=-7-24i. 724i=x2y2+2ixy-7-24i={{x}^{2}}-{{y}^{2}}+2ixy

Equating real and imaginary parts we get

x2y2=7{{x}^{2}}-{{y}^{2}}=-7

and 2xy=242xy=-24 (x2+y2)2=(x2y2)2+4x2y2{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}} =(7)2+(24)2=625={{\left( -7 \right)}^{2}}+{{\left( -24 \right)}^{2}}=625

x2+y2=25{{x}^{2}}+{{y}^{2}}=25

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

(x,y)=(3,4);(3,4)by(ii)\left( x,y \right)=\left( 3,-4 \right);\left( -3,4 \right)\,by\,\left( ii \right)

z0=±(34i){{z}_{0}}=\pm \left( 3-4i \right)

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

1+ωn+ω2n{3,whennisamultipleof30,whennisnotamultipleof31+{{\omega }^{n}}+{{\omega }^{2n}}\left\{ \begin{matrix} 3,\,when\,n\,is\,a\,multiple\,of\,3 \\ 0,when\,n\,is\,not\,a\,multiple\,of\,3 \\ \end{matrix} \right.

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

1+ωn+ω2n=1+ω3m+ω6m1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{3m}}+{{\omega }^{6m}} =1+1+1[Since,    ω3=1]=3=1+1+1\left[ Since,\;\;{{\omega }^{3}}=1 \right]=3

Case: II. n=3m+1or3m+2;mIn=3m+1\,or\,3m+2;m\in I

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

L.H.S=1ω3m+1+ω6m+2=1+ω+ω2=0L.H.S=1{{\omega }^{3m+1}}+{{\omega }^{6m+2}}=1+\omega +{{\omega }^{2}}=0

(b) Let n=3m+2n=3m+2 1+ω3m+2+ω6m+4=1+ω2+ω4=1+ω2+ω=0.1+{{\omega }^{3m+2}}+{{\omega }^{6m+4}}=1+{{\omega }^{2}}+{{\omega }^{4}}=1+{{\omega }^{2}}+\omega =0.

Illustration: Show that z3z+3=2\left| \frac{z-3}{z+3} \right|=2 represents a circle.

Sol. Consider z=x+iyz=x+iy

z3z+3=2x3+iyx+3+iy=2\left| \frac{z-3}{z+3} \right|=2\Rightarrow \left| \frac{x-3+iy}{x+3+iy} \right|=2 x3+iy2=22x+3+iy2{{\left| x-3+iy \right|}^{2}}={{2}^{2}}{{\left| x+3+iy \right|}^{2}}

or (x3)2+y2=4((x+3)2+y2){{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\left( {{\left( x+3 \right)}^{2}}+{{y}^{2}} \right) 3x2+3y2+30x+27=0\Rightarrow 3{{x}^{2}}+3{{y}^{2}}+30x+27=0

which represents a circle.

Illustration: If z1=z2=.=zn=1\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=…….=\left| {{z}_{n}} \right|=1

Prove that z1z2+.+zn=1z1+1z2+.+1zn\left| {{z}_{1}}{{z}_{2}}+…….+{{z}_{n}} \right|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+…….+\frac{1}{{{z}_{n}}} \right|

Sol. zj=1zjzˉj=1j=1,,n\left| {{z}_{j}} \right|=1\Rightarrow {{z}_{j}}{{\bar{z}}_{j}}=1\forall j=1,……,n (Since,  zzˉ=z2)\left( Since, \;z\bar{z}=\left| {{z}^{2}} \right| \right)

L.H.S.

z1z2+.+zn=1z1+1z2+.+1zn=\left| {{z}_{1}}{{z}_{2}}+…….+{{z}_{n}} \right|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+…….+\frac{1}{{{z}_{n}}} \right|= 1z1+1z2+1z3.+1zn\left| \overline{\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}…….+\frac{1}{{{z}_{n}}}} \right| =1z1+1z2+1z3.+1zn=R.H.S.=\left| \overline{\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}…….+\frac{1}{{{z}_{n}}}} \right|=R.H.S.

Illustration: If z1+z2=z1z2,\left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|, prove that argz1argz2=\arg {{z}_{1}}-\arg {{z}_{2}}= odd multiple of π2.\frac{\pi }{2}.

Sol. As we know z=z.zˉ.\left| z \right|=z.\bar{z}. Apply this formula and consider z=r(cosθ+isinθ).z=r\left( \cos \theta +i\,sin\theta \right). z1+z22=z1z22{{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}} (z1+z2)(zˉ1+zˉ2)=(z1z2)(zˉ1zˉ2)or\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 z1zˉ1+z2zˉ2+z2zˉ1+z1zˉ2=z1zˉ1+z2zˉ2z2zˉ1z1zˉ2{{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(z2zˉ1+z1zˉ2)=0;Re(z1zˉ2)=02\left( {{z}_{2}}{{{\bar{z}}}_{1}}+{{z}_{1}}{{{\bar{z}}}_{2}} \right)=0;{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)=0

Let z1=r1(cosθ1+isinθ1)andz2=r2(cosθ2+isinθ2);{{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 z1zˉ2=r1r2(cos(θ1θ2)+isin(θ1θ2)){{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(θ1θ2)=0(asRe(z1zˉ2)=0)\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=0\left( as\,{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)=0 \right) θ1θ2={{\theta }_{1}}-{{\theta }_{2}}= odd multiple of π2.\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+ ….+ anxn, then show that

(a) a0a2+a4+.=2n2cosnπ4{{a}_{0}}-{{a}_{2}}+{{a}_{4}}+….={{2}^{\frac{n}{2}}}\cos \frac{n\pi }{4}

(b) a1a3+a5+.=2n2sinnπ4{{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 + ….+ anin.

[2(cosπ4+isinπ4)]n{{\left[ \sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \right]}^{n}}= (a0 – a2 + a4 – …) + i (a1 – a3 + a5 – …)

2n/2(cosnπ4+isinnπ4){{2}^{n/2}}\left( \cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4} \right) = (a0 – a2 + a4 + …. ) + i (a1 – a3 + a5 + …)

Equating real and imaginary parts.

2n2cosnπ4=a0a2+a4+..{{2}^{\frac{n}{2}}}{{\cos }^{\frac{n\pi }{4}}}={{a}_{0}}-{{a}_{2}}+{{a}_{4}}+….. 2n2sinnπ4=a1a3+a5+..{{2}^{\frac{n}{2}}}{{\sin }^{\frac{n\pi }{4}}}={{a}_{1}}-{{a}_{3}}+{{a}_{5}}+…..

Example 9: Solve the equation zn1=z:nN{{z}^{n-1}}=\overline{z}:n\in N

Sol: Apply modulus on both side.

zn1=z;zn1=z=z{{z}^{n-1}}=\overline{z};\,\,\,\,\,\,\,\,|z{{|}^{n-1}}=|\overline{z}|=|z|

z=0orz=1Ifz=0thenz=0,\,\,|z|=0\,or\,|z|\,=1\,\,If\,|z|=0\,then\,z\,=\,0, Letz=1; then, zn=zz=1Let\left| z \right|=1;\text{ }then,\text{ }{{z}^{n}}=z\overline{z}=1

z=cos2mπn+sin2mπn:m=0,1,..,n1\,\,\,\,\,\,\,\,z=\cos \frac{2m\pi }{n}+\sin \frac{2m\pi }{n}:m=0,\,1,\,…..,\,n-1

Illustration: If z = x + iy and ω=1izzi\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

ω=1izzi=11iz=zi\left| \omega \right|=\left| \frac{1-iz}{z-i} \right|=1\Rightarrow |1-iz|=|z-i|or, |1 – ix + y| = |x + i(y – 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+17,1733+\sqrt{17},\sqrt{17}-3

(b) 6, 1

(c) 17,1\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.

JEE Complex numbers eg 1

 

Greatestdistance=3+12+42=3+17\,\,\,\,Greatest\,dis\tan ce\,=\,3+\sqrt{{{1}^{2}}+{{4}^{2}}}=3+\sqrt{17}

Least distance = 12+423=173\sqrt{{{1}^{2}}+{{4}^{2}}}-3=\sqrt{17}-3

Illustration: Find the equation of the circle for which arg (z62iz22i)=π/4.\left( \frac{z-6-2i}{z-2-2i} \right)=\pi /4.

Sol: arg(z62iz22i)=π/4\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 π4\frac{\pi }{4} at circumference.

JEE complex numbers eg 2

 

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

O=(4,4)andOA2=OM2+AM2=22\,\,O=(4,4)\,and\,O{{A}^{2}}=O{{M}^{2}}+A{{M}^{2}}=2\sqrt{2} Equation of required circle is

z44i=22|z-4-4i|=2\sqrt{2}

Illustration: if |z| > 3, prove that the least value of z+1zis83\left| z+\frac{1}{z} \right|\,is\,\frac{8}{3}

Sol: Here z+1z>z1z\left| z+\frac{1}{z} \right|\,\underline{>}\,|z|-\frac{1}{|z|} Now |z| > 3

1z<13or1z>13\,\,\,\,\frac{1}{|z|}\underline{<}\frac{1}{3}\,or\,-\frac{1}{|z|}\underline{>}-\frac{1}{3} (i)

Adding the two like inequalities

z1z>313=83|z|-\frac{1}{|z|}\underline{>}3-\frac{1}{3}=\frac{8}{3}

Hence from (i) and (ii), we get z+1z>83\left| z+\frac{1}{z} \right|\underline{>}\frac{8}{3} (ii)

Leastvalueis83\,\,\,Least\,value\,is\frac{8}{3}

Illustration: If z1, z2, z3 are non-zero complex numbers such that z1+z2+z3=0andz11+z21+z31=0{{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0\,and\,z_{1}^{-1}+z_{2}^{-1}+z_{3}^{-1}=0

then prove that the given points are the vertices of an equilateral triangle. Also show that

|z1| = |z2| = |z3|.

Sol: Use algebra to solve this problem.

Given z1 + z2 + z3 = 0, and from 2nd relation z2z3 + z3z1 + z1z2 = 0

z2z3=z1(z2+z3)=z1(z1)=z12