Complex Numbers

Introduction to Complex Numbers

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.

Illustration 5: If n is odd then (1 + i)Ion + (1 – i)Ion

Solution: (1 + i)ion + (1 – i)ion = [(1+i)2]5n + [(1 – i)2]5n

= (2i)5n + (–2i)5n = (2i)5n – (2i)5n = 0

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.

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

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

(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 + …. + an zn

Where a0, a1, ….. an 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

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

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

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

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

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

 

Q. \(\sqrt{3i}+\sqrt{-3i}\) =

Solution: \(\sqrt{3i}=\pm \sqrt{3}\left( \frac{1+i}{\sqrt{2}} \right)\) \(\sqrt{-3i}=\pm \sqrt{3}\left( \frac{1+i}{\sqrt{2}} \right)\)

by adding, \(\sqrt{3i}+\sqrt{-3i}=\pm \sqrt{3}\left( \frac{2}{\sqrt{2}} \right)=\pm \sqrt{6}\)

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

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

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

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

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

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

\(=\sqrt{{{x}^{2}}+{{y}^{2}}}\left[ \frac{2}{\sqrt{{{x}^{2}}+{{y}^{2}}}}+i\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right]\)

= |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

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

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

\(\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,z=\cos \left( \frac{2k\pi }{n} \right)+i\sin \left( \frac{2k\pi }{n} \right)\) [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 \(\omega =\cos \left( \frac{2\pi }{n} \right)+i\sin \left( \frac{2\pi }{n} \right)\)

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 = \(\frac{1-{{\omega }^{n}}}{1-\omega }\)

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

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

Also, note that

\(\frac{1}{x-1}+\frac{1}{x-\omega }….+\frac{1}{x-{{\omega }^{n-1}}}=\frac{n{{x}^{n-1}}}{{{x}^{n}}-1}\)

Cube roots of Unity

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

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

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

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

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 rth term of the given expression is

(r + 1) (r + ω) (r + ω2) = r3 + 1

Value of the given expression is

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

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

\(\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}.\)

Sol. We have, x3 – 3×2 + 3x + 7 = 0

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

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

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

\({{x}^{2}}-{{y}^{2}}=-7\)

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,

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

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

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

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

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

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

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.

\(\left| {{z}_{1}}{{z}_{2}}+…….+{{z}_{n}} \right|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+…….+\frac{1}{{{z}_{n}}} \right|=\) \(\left| \overline{\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}…….+\frac{1}{{{z}_{n}}}} \right|\) \(=\left| \overline{\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}…….+\frac{1}{{{z}_{n}}}} \right|=R.H.S.\)

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+ ….+ anxn, 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 + ….+ anin.

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

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

Equating real and imaginary parts.

\({{2}^{\frac{n}{2}}}{{\cos }^{\frac{n\pi }{4}}}={{a}_{0}}-{{a}_{2}}+{{a}_{4}}+…..\) \({{2}^{\frac{n}{2}}}{{\sin }^{\frac{n\pi }{4}}}={{a}_{1}}-{{a}_{3}}+{{a}_{5}}+…..\)

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

Sol: Apply modulus on both side.

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

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

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

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

\(\left| \omega \right|=\left| \frac{1-iz}{z-i} \right|=1\Rightarrow |1-iz|=|z-i|\)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) \(3+\sqrt{17},\sqrt{17}-3\)

(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

\(|z-4-4i|=2\sqrt{2}\)

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

\(|z|-\frac{1}{|z|}\underline{>}3-\frac{1}{3}=\frac{8}{3}\)

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

∴ \(\,\,\,Least\,value\,is\frac{8}{3}\)

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

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

∴ \(\,\,{{z}_{2}}{{z}_{3}}=-{{z}_{1}}({{z}_{2}}+{{z}_{3}})=-{{z}_{1}}(-{{z}_{1}})=z_{1}^{2}\)

∴ \(\,\,{{z}_{1}}^{3}={{z}_{1}}{{z}_{2}}{{z}_{3}}=z{{{}_{2}}^{3}}={{z}_{3}}^{3}\)

∴ \(\,\,|{{z}_{1}}{{|}^{3}}=|{{z}_{2}}{{|}^{3}}=|{{z}_{3}}{{|}^{3}}\)

Above shows that distance of origin from A, B, C is sample.

Origin is circumcentre, but z1 + z2 + z3 = 0

Implies that centroid is also at the origin so that the triangle must be equilateral.

Illustration: For constant c > 1, find all complex numbers z satisfying the equation z + c |z + 1| + i = 0

Sol: Solve this by putting z = x + iy.

Let z = x + iy

The equation z + c |z + 1| + i = 0 becomes

\(x+iy+c\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}+i=0\) \(or\,x+c\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}+i(y+1)=0\)

Equating real and imaginary parts, we get

y + 1 = 0 y = –1 (i)

and \(x+c\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=0:x<0\) (ii)

Solving (i) and (ii), we get

\(x+c\sqrt{{{(x+1)}^{2}}+1}=0\,or\,{{x}^{2}}={{c}^{2}}[{{(x+1)}^{2}}+1]\)

Or (c2 – 1)x2 + 2c2x + 2c2 = 0

If c = 1, then x = –1. Let c > 1; then,

\(x=\frac{-2{{c}^{2}}\pm \sqrt{4{{c}^{4}}-8{{c}^{2}}({{c}^{2}}-1)}}{2({{c}^{2}}-1)}=\frac{-{{c}^{2}}\pm c\sqrt{2-{{c}^{2}}}}{{{c}^{2}}-1}\)

As x is real and c > 1, we have: \(1<c\le \sqrt{2}\)

(Thus, for \(\sqrt{2}\), there is no solution). Since both values of x satisfy (ii), both values are admissible.

Illustration: find the sixth roots of z = 64i

Sol: \(Here\,i=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\) and sixth root of z

i.e. zr = z1/6.

\(z=64\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\) \({{z}_{r}}={{z}^{1/6}}\) \(=2\left( \cos \frac{2r\pi +\frac{\pi }{2}}{6}+\sin \frac{2r\pi +\frac{\pi }{2}}{6} \right)\)

Where r = 0, 1, 2, 3, 4, 5

The roots z0, z1, z3, z4, z5 are given by

\({{z}_{0}}=2\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right)\) \({{z}_{1}}=2\left( \cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12} \right)\) \({{z}_{2}}=2\left( \cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12} \right)\) \({{z}_{3}}=2\left( \cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12} \right)=-2\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right)\) \({{z}_{4}}=2\left( \cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12} \right)=-2\left( \cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12} \right)\) \({{z}_{5}}=2\left( \cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12} \right)=-2\left( \cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12} \right)\)

Illustration: Let three vertices A, B, C (taken in clock wise order) of an isosceles right angled triangle with right angle at C, be affixes of complex numbers z1, z2, z3 respectively. Show that (z1 – z2)2 = 2(z1 – z3) (z3 – z2).

Sol: Here \(\frac{{{z}_{2}}-{{z}_{3}}}{{{z}_{1}}-{{z}_{3}}}={{e}^{-i\pi /2}}.\)

Therefore solve it using algebra method.

Given CB = CA and angle \(\angle C=\frac{\pi }{2}\) \(\frac{{{z}_{2}}-{{z}_{3}}}{{{z}_{1}}-{{z}_{3}}}={{e}^{-i\pi /2}}\) or (z3 – z2)2 = i2(z1 – z3)2

(z3 – z2)2 = -(z1 – z3)2

Or

\(z_{3}^{2}+z_{2}^{2}-2{{z}_{2}}{{z}_{3}}+z_{1}^{2}+z_{3}^{2}-2{{z}_{1}}{{z}_{3}}=0\)

Add and subtract 2z1z2, we get

\(z_{1}^{2}+z_{2}^{2}-2{{z}_{1}}{{z}_{2}}+2z_{3}^{2}+2{{z}_{2}}{{z}_{3}}-2{{z}_{1}}{{z}_{3}}+2{{z}_{1}}{{z}_{2}}=0,\,or\) \({{({{z}_{1}}-{{z}_{2}})}^{2}}+2({{z}_{3}}-{{z}_{1}})({{z}_{3}}-{{z}_{2}})=0,\,or\,{{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})({{z}_{3}}-{{z}_{2}}).\)

Illustration: If A, B, C be the angles of triangle then

\(prove\,that\,\left| \begin{matrix} {{e}^{2iA}} & {{e}^{-iC}} & {{e}^{-iB}} \\ {{e}^{-iC}} & {{e}^{2iB}} & {{e}^{-iA}} \\ {{e}^{-iB}} & {{e}^{-iA}} & {{e}^{2iC}} \\ \end{matrix} \right|is\,purely\,real.\)

Sol: Here A+B+C = π, therefore epi = cos π + i sin π = –1. And by using properties of matrices we can solve this problem.

e-pi = –1 …. (i)

ei(B+C) = ei(π–A) = epi e–iA = –e–iA

ei(B+C)= –eiA …. (ii)

Take eiA, eiB and eiC common from R1, R2 and R3 respectively. Δ = ei(A+B+c)

\(\,\left| \begin{matrix} {{e}^{iA}} & {{e}^{-i(A+C)}} & {{e}^{-i(A+B)}} \\ {{e}^{-i(B+C)}} & {{e}^{iB}} & {{e}^{-i(B+A)}} \\ {{e}^{-i(B+C)}} & {{e}^{-i(C+A)}} & {{e}^{iC}} \\ \end{matrix} \right|=-1\left| \begin{matrix} {{e}^{iA}} & -{{e}^{iB}} & -{{e}^{i}} \\ -{{e}^{iA}} & {{e}^{iB}} & -{{e}^{iC}} \\ -{{e}^{iA}} & -{{e}^{iB}} & {{e}^{iC}} \\ \end{matrix} \right|,by\,(2)\)

Take eiA, eiB and eiC common from C1, C2 and C3 and again put ei(A+B+C)=e = –1

∴ \(\,\,\,\,\Delta =(-1)(-1)\left| \begin{matrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right|\)

Now make two zeros and expand Δ = –4 which is purely real.

Illustration: Show that all the roots of the equation \({{z}^{n}}cos{{q}_{0}}+{{z}^{n-1}}cos\text{ }{{q}_{1}}+{{z}^{n-2}}cos\,{{q}_{2}}+\ldots ..+z\,cos\text{ }{{q}_{n-1}}+cos\text{ }{{q}_{n}}=2\)

lie outside the circle \(|z|=\frac{1}{2}\) where q0, q1 etc are real.

Sol: By using triangle inequality.

Here \(\left| {{z}^{n}}cos\text{ }{{q}_{0}}+{{z}^{n-1}}cos\text{ }{{q}_{1}}+{{z}^{n-2}}\text{ }cos\text{ }{{q}_{2}}\text{ }+\text{ }\ldots .\text{ }+\text{ }z\text{ }cos\text{ }{{q}_{n-1}}+cos\text{ }{{q}_{n}} \right|=2\)

By triangle inequality.

2=\(|{{z}^{n}}cos{{\theta }_{n}}+{{2}^{n-1}}cos\,{{q}_{1}}+{{2}^{n-1}}\,\cos \,{{q}_{2}}+…+z\cos \,{{q}_{n-1}}+\cos \,{{q}_{n}}|\le |{{z}^{n}}cos{{q}_{n}}|+|{{z}^{n-1}}cos\,{{q}_{1}}|\)+

\(|{{z}^{n-2}}cos\,{{q}_{2}}|+….++|zcos\,{{q}_{n-1}}|+|cos\,{{q}_{n}}|=|{{z}_{n}}||cos\,{{q}_{n}}|+|{{z}^{n-1}}||cos\,{{q}_{n}}|+…+|z|cos\,{{q}_{n-1}}|+\) \(|cos\,{{q}_{n}}|\,\le \,|z{{|}^{n}}+|z{{|}^{n-1}}+….+|z|+1\) \(\left( Since,\;\,|cos\,{{q}_{1}}|\le 1\,and\,|{{z}^{n-1}}|=|z{{|}^{n+1}} \right)\) \(=\frac{1-|z{{|}^{n+1}}}{1-|z|}<\frac{1}{1-|z|}\)

∴ \(2<\frac{1}{1-|z|}\) \(sol\,1-|z|is\,positive\,and\,1-|z|\,<\frac{1}{2}\)

∴ \(\,|z|>1-\frac{1}{2}=\frac{1}{2}\)

∴ \(\,\,\,All\,z\,satisfying\,(i)\,lie\,outside\,the\,circle\,|z|=\frac{1}{2}\)

Illustration: If \(z+\frac{1}{z}=2\cos \theta ,prove\,that\,\left| \frac{{{z}^{2n}}-1}{{{z}^{2n}}+1} \right|=|tan\,nq|\)

Sol: By using formula of roots of quadratic equation, we can solve this problem.

Here z + 1/z = 2 cos θ

∴ \(\,{{z}^{2}}-2\cos \,\theta .\,z+1=0\)

∴ \(\,z=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}=\cos \theta \pm i\sin \theta\)

Taking positive sign, z = cos Ɵ – i sin Ɵ

∴ \(\,\frac{1}{z}={{(cos\,\theta \,+\,i\,sin\,\theta )}^{-1}}=cos\,\theta \,-\,i\,sin\,\theta\)

∴ \(\,\frac{{{z}^{2n}}-1}{{{z}^{2n}}+1}=\frac{{{z}^{n}}-\frac{1}{{{z}^{n}}}}{{{z}^{n}}+\frac{1}{{{z}^{n}}}}=\frac{{{(cos\theta +isin\theta )}^{n}}-{{(cos\theta -isin\theta )}^{n}}}{{{(cos\theta +isin\theta )}^{n}}+{{(cos\theta -isin\theta )}^{n}}}\) \(=\frac{\cos n\theta +i\sin \theta -(cosn\theta -isin\theta )}{\cos n\theta +i\sin \theta +(cosn\theta -isinn\theta }=\frac{2i\,\sin n\theta }{2\cos n\theta }=i\tan n\theta\). Taking negative sign,

Similarly we get \(\frac{{{z}^{2n}}-1}{{{z}^{2n}}+1}=\frac{-2i\sin \,n\theta }{2\cos n\theta }=-i\tan \,n\theta\)

∴ \(\,\left| \frac{{{z}^{2n}}-1}{{{z}^{2n}}+1} \right|=|\pm i\,tan\,nq|=|tan\,nq|,\)

Illustration: Find the complex number z which satisfies the condition |z – 2 + 2i| = 1 and has the least absolute value.

Sol: Here z – 2 + 2i = cos Ɵ + i sin Ɵ, therefore by obtaining modulus of z we can solve above problem.

|z – 2 + 2i| = 1

\(\Rightarrow \,z-2+2i=\cos \,\theta +i\,\sin \,\theta\)

Where Ɵ is some real numbers.

\(\Rightarrow \,z=(2+cos\theta )+(sin\theta -2)i\) \(\Rightarrow \,|z|={{[{{(2+cos\,\theta )}^{2}}+{{(sin\theta -2)}^{2}}]}^{1/2}}\) \(={{[8+co{{s}^{2}}\theta +si{{n}^{2}}\theta +4(cos\theta -\sin \theta )]}^{1/2}}\) \(=\,{{\left[ 9+4\sqrt{2}\cos \left( \theta +\frac{\pi }{4} \right) \right]}^{1/2}}\)

|z| will be least if cos (Ɵ + π/4) is least, that is, if cos (Ɵ + π/4) = -1 or Ɵ =\(\frac{3\pi }{4}\). Thus, least value of |z| is

\({{\left( 9-4\sqrt{2} \right)}^{1/2}}for\,z=\left( 2-\frac{1}{\sqrt{2}} \right)+i\left( \frac{1}{\sqrt{2}}-2 \right)\)

 


Practise This Question

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+n2) is equal to