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 a Real Number and an Imaginary Number .
Table of Content 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 = √-1 we have i2 = –1 , i3 = –i, i4 = i.
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 = (i)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
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) + (-2+2i)
= 1 + 1 + i + 2i -2+2i = 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.
Check: JEE Previous Year Solved Problems on Complex Numbers
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.Read More
Properties of Conjugate of complex Number (i)
\(\begin{array}{l}{{z}_{1}}={{z}_{2}}\Leftrightarrow {{\overline{z}}_{1}}={{\overline{z}}_{2}}\end{array} \)
(ii)
\(\begin{array}{l}\overline{(\bar{z})}=z\end{array} \)
(iii)
\(\begin{array}{l}z+\overline{z}=2\,{Re}(z)\end{array} \)
(iv)
\(\begin{array}{l}z-\overline{z}=2i\,{Im}\,(z)\end{array} \)
(v)
\(\begin{array}{l}z=\overline{z}\Leftrightarrow z\end{array} \)
is purely real
(vi)
\(\begin{array}{l}z+\overline{z}=0\Leftrightarrow z\end{array} \)
is purely imaginary.
(vii)
\(\begin{array}{l}z\overline{z}={{[Re\,(z)]}^{2}}+{{[Im(z)]}^{2}}\end{array} \)
(viii)
\(\begin{array}{l}\overline{{{z}_{1}}+{{z}_{2}}}={{\overline{z}}_{1}}+{{\overline{z}}_{2}}\end{array} \)
(ix)
\(\begin{array}{l}\overline{{{z}_{1}}-{{z}_{2}}}={{\overline{z}}_{1}}-{{\overline{z}}_{2}}\end{array} \)
(x)
\(\begin{array}{l}\overline{{{z}_{1}}{{z}_{2}}}={{\overline{z}}_{_{1}}}{{\overline{z}}_{_{2}}}\end{array} \)
(xi)
\(\begin{array}{l}\overline{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}=\frac{{{\overline{z}}_{1}}}{{{z}_{2}}}\end{array} \)
if z2 ≠ 0
(xii) If P(z) = a0 + a1 z + a2 z2 + …. + an zn
Where a0 , a1 , ….. an and z are complex number, then
\(\begin{array}{l}\overline{P(z)}={{\overline{a}}_{0}}+{{\overline{a}}_{1}}(\overline{z})+{{\overline{a}}_{2}}{{(\overline{z})}^{2}}+….+{{\overline{a}}_{n}}{{(\overline{z})}^{n}}\end{array} \)
=
\(\begin{array}{l}\overline{P}(\overline{z})\end{array} \)
Where
\(\begin{array}{l}\overline{P}(z)={{\overline{a}}_{0}}+{{\overline{a}}_{1}}z+{{\overline{a}}_{2}}{{z}^{2}}+….+{{\overline{a}}_{n}}{{z}^{n}}\end{array} \)
(xiii) If R(z) =
\(\begin{array}{l}\frac{P(z)}{Q(z)}\end{array} \)
where P (z) and Q (z) are polynomials in z, and Q(z) ≠ 0, then
\(\begin{array}{l}\overline{R\,(z)}=\frac{\overline{P}(\overline{z})}{\overline{Q}(\overline{z})}\end{array} \)
Conjugates of Complex Numbers Video Lesson
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
Read More
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 +…+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 ||
Square Root of a Complex Number Let z = x + iy then
\(\begin{array}{l}\sqrt{x+iy}=\left\{ \begin{matrix} \pm \left[ \sqrt{\frac{|z|+x}{2}}+i\sqrt{\frac{|z|-x}{2}} \right]if\,y\,>\,0 \\ \pm \left[ \sqrt{\frac{|z|+x}{2}}-i\sqrt{\frac{|z|-x}{2}} \right]if\,y\,<\,0 \\ \end{matrix} \right.\end{array} \)
Where |x| =
\(\begin{array}{l}\sqrt{{{x}^{2}}+{{y}^{2}}}\,\end{array} \)
NOTE:
(i)
\(\begin{array}{l}\sqrt{x+iy}+\sqrt{x-iy}=\sqrt{2|z|+2x}\end{array} \)
(ii)
\(\begin{array}{l}\sqrt{x+iy}-\sqrt{x-iy}=i\sqrt{2|z|-2x}\end{array} \)
(iii)
\(\begin{array}{l}\sqrt{i}=\pm \left( \frac{1+i}{\sqrt{2}} \right)\,and\,\sqrt{-i}=\pm \left( \frac{1-i}{\sqrt{2}} \right)\end{array} \)
Modulus and Argument of a Complex Number Let z = x + iy = (x, y) for all x, y
\(\begin{array}{l}\in\end{array} \)
R and i = \(\begin{array}{l}\sqrt{-1}\end{array} \)
The length OP is called modulus of the complex number z denoted by |z|,
i.e. OP = r = |z| =
\(\begin{array}{l}\sqrt{({{x}^{2}}+{{y}^{2}})}\end{array} \)
and if (x, y) ≠ (0, 0), then θ is called the argument or amplitude of z,
i.e. θ =
\(\begin{array}{l}{{\tan }^{-1}}\left( \frac{y}{x} \right)\end{array} \)
[angle made by OP with positive X-axis]
or arg (z) =
\(\begin{array}{l}{{\tan }^{-1}}\left( y/x \right)\end{array} \)
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
\(\begin{array}{l}\in\end{array} \)
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 =
\(\begin{array}{l}{{\tan }^{-1}}\left| \frac{y}{x} \right|\end{array} \)
.
Illustration 5 . Find the arguments of z1 = 5 + 5i, z2 = –4 + 4i, z3 = –3 – 3i and z4 = 2 – 2i, where
\(\begin{array}{l}i=\sqrt{-1}\end{array} \)
.
Solution: Since, z1 , z2 , z3 and z4 lies in I, II, III and IV quadrants respectively. The arguments are given by
arg (z1 ) =
\(\begin{array}{l}{{\tan }^{-1}}\left( \frac{5}{5} \right)={{\tan }^{-1}}1=\pi /4\end{array} \)
arg (z2 ) =
\(\begin{array}{l}\pi -{{\tan }^{-1}}\left| \frac{4}{-4} \right|=\pi -{{\tan }^{-1}}1=\pi -\frac{\pi}{4}=\frac{3\pi }{4}\end{array} \)
arg (z3 ) =
\(\begin{array}{l}\pi -{{\tan }^{-1}}\left| \frac{-3}{-3} \right|=\pi +{{\tan }^{-1}}1=\pi +\frac{\pi}{4}=\frac{5\pi }{4}\end{array} \)
And arg (z4 ) =
\(\begin{array}{l}2\pi -{{\tan }^{-1}}\left| \frac{-2}{2} \right|=2\pi +{{\tan }^{-1}}1=2\pi -\frac{\pi}{4}=\frac{7\pi }{4}\end{array} \)
Principal value of the Argument The value θ of the argument which satisfies the inequality
\(\begin{array}{l}-\pi <\theta \le \pi\end{array} \)
is called the principal value of the argument.
If x = x + iy = ( x, y),
\(\begin{array}{l}\forall\end{array} \)
x, y \(\begin{array}{l}\in\end{array} \)
R and \(\begin{array}{l}i=\sqrt{-1}\end{array} \)
, then
Arg(z) =
\(\begin{array}{l}{{\tan }^{-1}}\left( \frac{y}{x} \right)\end{array} \)
always gives the principal value. It depends on the quadrant in which the point (x, y) lies.
(i) (x, y)
\(\begin{array}{l}\in\end{array} \)
first quadrant x > 0, y > 0.
The principal value of arg (z) =
\(\begin{array}{l}\theta ={{\tan }^{-1}}\left( \frac{y}{x} \right)\end{array} \)
It is an acute angle and positive.
(ii) (x, y)
\(\begin{array}{l}\in\end{array} \)
second quadrant x < 0, y > 0.
The principal value of arg (z) = θ =
\(\begin{array}{l}\pi -{{\tan }^{-1}}\left( \frac{y}{|x|}\right)\end{array} \)
. It is an obtuse angle and positive.
(iii) (x, y)
\(\begin{array}{l}\in\end{array} \)
third quadrant x < 0, y < 0.
The principal value of arg (z) = θ =
\(\begin{array}{l}-\pi +{{\tan }^{-1}}\left( \frac{y}{x}\right)\end{array} \)
It is an obtuse angle and negative.
(iv) (x, y)
\(\begin{array}{l}\in\end{array} \)
fourth quadrant x > 0, y < 0.
The principal value of arg (z) = θ =
\(\begin{array}{l}-{{\tan }^{-1}}\left( \frac{|y|}{x}\right)\end{array} \)
It is an acute angle and negative.
Polar Form of a Complex Number We have, z = x + iy
\(\begin{array}{l}=\sqrt{{{x}^{2}}+{{y}^{2}}}\left[ \frac{2}{\sqrt{{{x}^{2}}+{{y}^{2}}}}+i\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right]\end{array} \)
= |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 eiƟ = 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| eiƟ
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 the 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
\(\begin{array}{l}\cos \left[ \frac{p}{q}(2k\pi +\theta ) \right]+i\,sin\left[ \frac{p}{q}(2k\pi +\theta ) \right]\end{array} \)
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
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,z=\cos \left( \frac{2k\pi }{n} \right)+i\sin \left( \frac{2k\pi }{n} \right)\end{array} \)
[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 the case of 3, we may take -1, 0 and 1 and in the case of 4, we may take – 1, 0, 1 and 2 or -2, -1, 0 and 1.
Notation
\(\begin{array}{l}\omega =\cos \left( \frac{2\pi }{n} \right)+i\sin \left( \frac{2\pi }{n} \right)\end{array} \)
By using 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 =
\(\begin{array}{l}\frac{1-{{\omega }^{n}}}{1-\omega }\end{array} \)
But ωn = 1 as ω is a nth root of unity.
∴
\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+\omega +…+{{\omega }^{n-1}}=0\end{array} \)
Also, note that
\(\begin{array}{l}\frac{1}{x-1}+\frac{1}{x-\omega }….+\frac{1}{x-{{\omega }^{n-1}}}=\frac{n{{x}^{n-1}}}{{{x}^{n}}-1}\end{array} \)
nth Root of Unity Video Lesson
Cube roots of Unity
Cube roots of unity are given by 1, ω, ω2 , where
\(\begin{array}{l}\omega =\cos \left( \frac{2\pi }{3} \right)+i\sin \left( \frac{2\pi }{3} \right)=\frac{-1+\sqrt{3i}}{2}and\,{{\omega }^{2}}=\frac{-1-\sqrt{3i}}{2}\end{array} \)
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 , a1/3 ω, 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 .
Cube Roots of Unity Video Lesson
Logarithm of a Complex Number Loge (x + iy) = loge (|z|eiƟ )
= loge |z| + loge eiƟ
= loge |z| + iƟ
=
\(\begin{array}{l}{{\log }_{e}}\sqrt{({{x}^{2}}+{{y}^{2}})}+i\arg (z)\end{array} \)
∴
\(\begin{array}{l}\,\,\,\,\,\,\,\,\,{{\log }_{e}}(z)=lo{{g}_{e}}|z|+iarg(z)\end{array} \)
Problems on Complex Numbers Illustration 1: The number of solutions of
\(\begin{array}{l}{{z}^{3}}+\overline{z}=0\end{array} \)
is
(a) 2 (b) 3 (c) 4 (d) 5
Ans. (d)
Solution
\(\begin{array}{l}{{z}^{3}}+\overline{z}=0\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{{z}^{3}}=-\overline{z}\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,|z{{|}^{3}}=|-\overline{z}|\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,|z{{|}^{3}}=|z|\end{array} \)
\(\begin{array}{l}|z|\,(|z|-1)\,(|z|+1)=0\end{array} \)
⇒ |z|=0 or |z|=1 [Since, |z|+1>0]
If |z| = 0, then z = 0
If
\(\begin{array}{l}|z|=1\,\,\,we\,get\,|z{{|}^{2}}=1\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,z\,\overline{z}=\,1\end{array} \)
Thus,
\(\begin{array}{l}{{z}^{3}}+\overline{z}=0\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{3}}+1/z=0\,\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,{{z}^{4}}+1=0\end{array} \)
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 2: 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)
\(\begin{array}{l}\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}+n\end{array} \)
(b)
\(\begin{array}{l}\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}-n\end{array} \)
(c)
\(\begin{array}{l}\frac{1}{4}n{{(n+1)}^{2}}-n\end{array} \)
(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
\(\begin{array}{l}\sum\limits_{r=1}^{n}{({{r}^{3}}+1)}=\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}+n\end{array} \)
Illustration 3: Find the real part of
\(\begin{array}{l}{{(1-i)}^{-i}}\end{array} \)
Sol: Let z =
\(\begin{array}{l}{{(1-i)}^{-i}}\end{array} \)
. Taking log on both sides, we have \(\begin{array}{l}log\,z=-i\,lo{{g}_{e}}(1-i)\end{array} \)
=
\(\begin{array}{l}-i{{\log }_{e}}\sqrt{2}\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right)\end{array} \)
=
\(\begin{array}{l}-i{{\log }_{e}}(\sqrt{2}{{e}^{-i}}^{(\pi /4)})\end{array} \)
=
\(\begin{array}{l}-i\left[ \frac{1}{2}{{\log }_{e}}2+{{\log }_{e}}^{-i\pi /4} \right]\end{array} \)
=
\(\begin{array}{l}-i\left[ \frac{1}{2}{{\log }_{e}}2-\frac{i\pi }{4} \right]\end{array} \)
=
\(\begin{array}{l}-\frac{i}{2}{{\log }_{e}}2-\frac{\pi }{4}\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,z={{e}^{-\pi /4}}\,{{e}^{-i(log\,2)/2}}\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{Re}(z)={{e}^{-\pi /4}}\cos \left( \frac{1}{2}\log 2 \right)\end{array} \)
Illustration 4: If α, β and γ are the roots of x3 – 3x2 + 3x + 7 = 0,(ω is the cube roots of unity) find the value of
\(\begin{array}{l}\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}.\end{array} \)
Sol. We have, x3 – 3x2 + 3x + 7 = 0
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(x-1)}^{3}}+8=0\end{array} \)
,
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(x-1)}^{3}}+{{2}^{3}}=0\end{array} \)
,
\(\begin{array}{l}\Rightarrow \,\,\,(x-1+2)(x-1+2\omega )(x-1+2{{\omega }^{2}})=0\end{array} \)
,
\(\begin{array}{l}\Rightarrow \,\,\,(x+1)(x-1+2\omega )(x-1+2{{\omega }^{2}})=0\end{array} \)
,
∴
\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-1,1-2\omega ,1-2{{\omega }^{2}}\end{array} \)
,
\(\begin{array}{l}\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha =-1,\beta =1-2\omega ,\gamma=1-2{{\omega}^{2}}\end{array} \)
Then,
\(\begin{array}{l}\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}=\frac{-2}{-2\omega }+\frac{-2\omega }{-2{{\omega }^{2}}}+\frac{-2{{\omega }^{2}}}{-2}\end{array} \)
\(\begin{array}{l}=\frac{1}{\omega }+\frac{1}{\omega }+{{\omega }^{2}}={{\omega }^{2}}+{{\omega }^{2}}+{{\omega }^{2}}=3{{\omega }^{2}}\end{array} \)
Illustration 5:
\(\begin{array}{l}If\,\,{{z}_{1}}\,\,and\,{{z}_{2}}\,are\,\,1-i,-2+4i\end{array} \)
respectively. Find \(\begin{array}{l}{Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{z}_{1}}} \right).\end{array} \)
Sol.
\(\begin{array}{l}\frac{{{z}_{1}}{{z}_{2}}}{{{z}_{1}}}=\frac{\left( 1-i \right)\left( -2+4i \right)}{1+i}=\frac{-2+2i+4i+4}{1+i}\end{array} \)
\(\begin{array}{l}=\frac{2+6i}{1+i}\times \frac{1-i}{1-i}=\frac{2+6i-2i+6}{2}=4+2i\end{array} \)
∴
\(\begin{array}{l}\,\,\,\,\,{Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}} \right)=2.\end{array} \)
Illustration 6: Find the square root of
\(\begin{array}{l}z=-7-24i.\end{array} \)
Sol. Consider
\(\begin{array}{l}{{z}_{0}}=x+iy\end{array} \)
be a square root then \(\begin{array}{l}{{z}_{0}}^{2}=-7-24i.\end{array} \)
,
or
\(\begin{array}{l}-7-24i={{x}^{2}}-{{y}^{2}}+2ixy\end{array} \)
Equating real and imaginary parts we get
\(\begin{array}{l}{{x}^{2}}-{{y}^{2}}=-7\end{array} \)
and
\(\begin{array}{l}2xy=-24\end{array} \)
,
\(\begin{array}{l}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}}\end{array} \)
\(\begin{array}{l}={{\left( -7 \right)}^{2}}+{{\left( -24 \right)}^{2}}=625\end{array} \)
,
∴
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}=25\end{array} \)
,
Solving (i) and (iii), we get,
\(\begin{array}{l}\left( x,y \right)=\left( 3,-4 \right);\left( -3,4 \right)\,by\,\left( ii \right)\end{array} \)
,
∴
\(\begin{array}{l}{{z}_{0}}=\pm \left( 3-4i \right)\end{array} \)
Illustration 7: If n is a positive integer and
\(\begin{array}{l}\omega\end{array} \)
be an imaginary cube root of unity, prove that
\(\begin{array}{l}1+{{\omega }^{n}}+{{\omega }^{2n}}=\left\{ \begin{matrix} 3,\,when\,n\,is\,a\,multiple\,of\,3 \\ 0,when\,n\,is\,not\,a\,multiple\,of\,3 \\ \end{matrix} \right.\end{array} \)
Sol. Case: I.
\(\begin{array}{l}n=3m;m\in I\end{array} \)
∴
\(\begin{array}{l}1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{3m}}+{{\omega }^{6m}}\end{array} \)
,
\(\begin{array}{l}=1+1+1\left[ Since,\;\;{{\omega }^{3}}=1 \right]=3\end{array} \)
Case: II.
\(\begin{array}{l}n=3m+1\,or\,3m+2;m\in I\end{array} \)
(a) Let
\(\begin{array}{l}n=3m+1\end{array} \)
∴
\(\begin{array}{l}L.H.S=1+{{\omega }^{3m+1}}+{{\omega }^{6m+2}}=1+\omega +{{\omega }^{2}}=0\end{array} \)
(b) Let
\(\begin{array}{l}n=3m+2\end{array} \)
,
\(\begin{array}{l}1+{{\omega }^{3m+2}}+{{\omega }^{6m+4}}=1+{{\omega }^{2}}+{{\omega }^{4}}=1+{{\omega }^{2}}+\omega =0.\end{array} \)
Illustration 8: Show that
\(\begin{array}{l}\left| \frac{z-3}{z+3} \right|=2\end{array} \)
represents a circle.
Sol. Consider
\(\begin{array}{l}z=x+iy\end{array} \)
∴
\(\begin{array}{l}\left| \frac{z-3}{z+3} \right|=2\Rightarrow \left| \frac{x-3+iy}{x+3+iy} \right|=2\end{array} \)
,
\(\begin{array}{l}{{\left| x-3+iy \right|}^{2}}={{2}^{2}}{{\left| x+3+iy \right|}^{2}}\end{array} \)
or
\(\begin{array}{l}{{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\left( {{\left( x+3 \right)}^{2}}+{{y}^{2}} \right)\end{array} \)
\(\begin{array}{l}\Rightarrow 3{{x}^{2}}+3{{y}^{2}}+30x+27=0\end{array} \)
which represents a circle.
Illustration 9: If
\(\begin{array}{l}\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=…….=\left| {{z}_{n}} \right|=1\end{array} \)
Prove that
\(\begin{array}{l}\left| {{z}_{1}}{{z}_{2}}+…….+{{z}_{n}} \right|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+…….+\frac{1}{{{z}_{n}}} \right|\end{array} \)
Sol.
\(\begin{array}{l}\left| {{z}_{j}} \right|=1\Rightarrow {{z}_{j}}{{\bar{z}}_{j}}=1\forall j=1,……,n\end{array} \)
\(\begin{array}{l}\left( Since, \;z\bar{z}=\left| {{z}^{2}} \right| \right)\end{array} \)
L.H.S.
\(\begin{array}{l}\left| {{z}_{1}}{{z}_{2}}+…….+{{z}_{n}} \right|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+…….+\frac{1}{{{z}_{n}}} \right|=\end{array} \)
\(\begin{array}{l}\left| \overline{\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}…….+\frac{1}{{{z}_{n}}}} \right|\end{array} \)
\(\begin{array}{l}=\left| \overline{\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}…….+\frac{1}{{{z}_{n}}}} \right|=R.H.S.\end{array} \)
Illustration 10: If
\(\begin{array}{l}\left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|,\end{array} \)
prove that \(\begin{array}{l}\arg {{z}_{1}}-\arg {{z}_{2}}=\end{array} \)
odd multiple of \(\begin{array}{l}\frac{\pi }{2}.\end{array} \)
Sol. As we know
\(\begin{array}{l}\left| z \right|=z.\bar{z}.\end{array} \)
Apply this formula and consider \(\begin{array}{l}z=r\left( \cos \theta +i\,sin\theta \right).\end{array} \)
,
\(\begin{array}{l}{{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}\end{array} \)
\(\begin{array}{l}\Rightarrow \left( {{z}_{1}}+{{z}_{2}} \right)\left( {{{\bar{z}}}_{1}}+{{{\bar{z}}}_{2}} \right)=\left( {{z}_{1}}-{{z}_{2}} \right)\left( {{{\bar{z}}}_{1}}-{{{\bar{z}}}_{2}} \right)\,\,or\end{array} \)
\(\begin{array}{l}{{z}_{1}}{{\bar{z}}_{1}}+{{z}_{2}}{{\bar{z}}_{2}}+{{z}_{2}}{{\bar{z}}_{1}}+{{z}_{1}}{{\bar{z}}_{2}}={{z}_{1}}{{\bar{z}}_{1}}+{{z}_{2}}{{\bar{z}}_{2}}-{{z}_{2}}{{\bar{z}}_{1}}-{{z}_{1}}{{\bar{z}}_{2}}\end{array} \)
or
\(\begin{array}{l}2\left( {{z}_{2}}{{{\bar{z}}}_{1}}+{{z}_{1}}{{{\bar{z}}}_{2}} \right)=0;{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)=0\end{array} \)
Let
\(\begin{array}{l}{{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\,\sin {{\theta }_{1}} \right)\,and\,{{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right);\end{array} \)
then
\(\begin{array}{l}{{z}_{1}}{{\bar{z}}_{2}}={{r}_{1}}{{r}_{2}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right)\end{array} \)
∴
\(\begin{array}{l}\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=0\left( as\,{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)=0 \right)\end{array} \)
,
\(\begin{array}{l}{{\theta }_{1}}-{{\theta }_{2}}=\end{array} \)
odd multiple of \(\begin{array}{l}\frac{\pi }{2}.\end{array} \)
Illustration 11: 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) =5+3= 8
Illustration 12: If (1 + x)n = a0 + a1 x + a2 x2 + ….+ an xn , then show that
(a)
\(\begin{array}{l}{{a}_{0}}-{{a}_{2}}+{{a}_{4}}+….={{2}^{\frac{n}{2}}}\cos \frac{n\pi }{4}\end{array} \)
(b)
\(\begin{array}{l}{{a}_{1}}-{{a}_{3}}+{{a}_{5}}+….={{2}^{\frac{n}{2}}}\sin \frac{n\pi }{4}\end{array} \)
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 + a1 i + a2 i2 + ….+ an in .
\(\begin{array}{l}{{\left[ \sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \right]}^{n}}\end{array} \)
= (a0 – a2 + a4 – …) + i (a1 – a3 + a5 – …)
\(\begin{array}{l}{{2}^{n/2}}\left( \cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4} \right)\end{array} \)
= (a0 – a2 + a4 + …. ) + i (a1 – a3 + a5 + …)
Equating real and imaginary parts.
\(\begin{array}{l}{{2}^{\frac{n}{2}}}{{\cos }^{\frac{n\pi }{4}}}={{a}_{0}}-{{a}_{2}}+{{a}_{4}}+…..\end{array} \)
,
\(\begin{array}{l}{{2}^{\frac{n}{2}}}{{\sin }^{\frac{n\pi }{4}}}={{a}_{1}}-{{a}_{3}}+{{a}_{5}}+…..\end{array} \)
Therefore, (a) and (b) are proved.
Illustration 13: Solve the equation
\(\begin{array}{l}{{z}^{n-1}}=\overline{z}:n\in N\end{array} \)
Sol: Apply modulus on both side.
\(\begin{array}{l}{{z}^{n-1}}=\overline{z};\,\,\,\,\,\,\,\,|z{{|}^{n-1}}=|\overline{z}|=|z|\end{array} \)
∴
\(\begin{array}{l}\,\,|z|=0\,or\,|z|\,=1\,\,If\,|z|=0\,then\,z\,=\,0,\end{array} \)
,
\(\begin{array}{l}Let\left| z \right|=1;\text{ }then,\text{ }{{z}^{n}}=z\overline{z}=1\end{array} \)
∴
\(\begin{array}{l}\,\,\,\,\,\,\,\,z=\cos \frac{2m\pi }{n}+\sin \frac{2m\pi }{n}:m=0,\,1,\,…..,\,n-1\end{array} \)
Illustration 14: If z = x + iy and
\(\begin{array}{l}\omega =\frac{1-iz}{z-i}\end{array} \)
with |ω| = 1, show that, z ; lies on the real axis.
Sol: Substitute value of ω in |ω| = 1
\(\begin{array}{l}\left| \omega \right|=\left| \frac{1-iz}{z-i} \right|=1\Rightarrow |1-iz|=|z-i|\end{array} \)
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 15: 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)
\(\begin{array}{l}3+\sqrt{17},\sqrt{17}-3\end{array} \)
(b) 6, 1
(c)
\(\begin{array}{l}\sqrt{17},1\end{array} \)
(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.
∴
\(\begin{array}{l}\,\,\,\,Greatest\,dis\tan ce\,=\,3+\sqrt{{{1}^{2}}+{{4}^{2}}}=3+\sqrt{17}\end{array} \)
Least distance =
\(\begin{array}{l}\sqrt{{{1}^{2}}+{{4}^{2}}}-3=\sqrt{17}-3\end{array} \)
Illustration 16: Find the equation of the circle for which arg
\(\begin{array}{l}\left( \frac{z-6-2i}{z-2-2i} \right)=\pi /4.\end{array} \)
Sol:
\(\begin{array}{l}\arg \left( \frac{z-6-2i}{z-2-2i} \right)=\pi /4\end{array} \)
represent a major arc of circle of which Line joining (6, 2)) and (2, 2) is a chord that subtends an angle
\(\begin{array}{l}\frac{\pi }{4}\end{array} \)
at circumference.
Clearly AB is parallel to real (x) axis, M is mid-point, M ≡ (4, 2), OM = AM = 2
∴
\(\begin{array}{l}\,\,O=(4,4)\,and\,O{{A}^{2}}=O{{M}^{2}}+A{{M}^{2}}=2\sqrt{2}\end{array} \)
Equation of required circle is
\(\begin{array}{l}|z-4-4i|=2\sqrt{2}\end{array} \)
Illustration 17 : if |z| > 3, prove that the least value of
\(\begin{array}{l}\left| z+\frac{1}{z} \right|\,is\,\frac{8}{3}\end{array} \)
Sol: Here
\(\begin{array}{l}\left| z+\frac{1}{z} \right|\,\underline{>}\,|z|-\frac{1}{|z|}\end{array} \)
Now |z| > 3
∴
\(\begin{array}{l}\,\,\,\,\frac{1}{|z|}\underline{<}\frac{1}{3}\,or\,-\frac{1}{|z|}\underline{>}-\frac{1}{3}\end{array} \)
……(i)
Adding the two like inequalities
\(\begin{array}{l}|z|-\frac{1}{|z|}\underline{>}3-\frac{1}{3}=\frac{8}{3}\end{array} \)
Hence from (i) and (ii), we get
\(\begin{array}{l}\left| z+\frac{1}{z} \right|\underline{>}\frac{8}{3}\end{array} \)
….. (ii)
∴ Least value is 8/3.
Illustration 18: If z1 , z2 , z3 are non-zero complex numbers such that
\(\begin{array}{l}{{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0\,and\,z_{1}^{-1}+z_{2}^{-1}+z_{3}^{-1}=0\end{array} \)
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 z2 z3 + z3 z1 + z1 z2 = 0
∴
\(\begin{array}{l}\,\,{{z}_{2}}{{z}_{3}}=-{{z}_{1}}({{z}_{2}}+{{z}_{3}})=-{{z}_{1}}(-{{z}_{1}})=z_{1}^{2}\end{array} \)
∴
\(\begin{array}{l}\,\,{{z}_{1}}^{3}={{z}_{1}}{{z}_{2}}{{z}_{3}}=z{{{}_{2}}^{3}}={{z}_{3}}^{3}\end{array} \)
∴
\(\begin{array}{l}\,\,|{{z}_{1}}{{|}^{3}}=|{{z}_{2}}{{|}^{3}}=|{{z}_{3}}{{|}^{3}}\end{array} \)
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 19: 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
\(\begin{array}{l}x+iy+c\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}+i=0\end{array} \)
\(\begin{array}{l}or\,x+c\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}+i(y+1)=0\end{array} \)
Equating real and imaginary parts, we get
y + 1 = 0 y = –1 (i)
and
\(\begin{array}{l}x+c\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=0:x<0\end{array} \)
(ii)
Solving (i) and (ii), we get
\(\begin{array}{l}x+c\sqrt{{{(x+1)}^{2}}+1}=0\,or\,{{x}^{2}}={{c}^{2}}[{{(x+1)}^{2}}+1]\end{array} \)
Or (c2 – 1)x2 + 2c2 x + 2c2 = 0
If c = 1, then x = –1. Let c > 1; then,
\(\begin{array}{l}x=\frac{-2{{c}^{2}}\pm \sqrt{4{{c}^{4}}-8{{c}^{2}}({{c}^{2}}-1)}}{2({{c}^{2}}-1)}=\frac{-{{c}^{2}}\pm c\sqrt{2-{{c}^{2}}}}{{{c}^{2}}-1}\end{array} \)
As x is real and c > 1, we have:
\(\begin{array}{l}1<c\le \sqrt{2}\end{array} \)
(Thus, for
\(\begin{array}{l}\sqrt{2}\end{array} \)
, there is no solution). Since both values of x satisfy (ii), both values are admissible.
Illustration 20: find the sixth roots of z = 64i
Sol:
\(\begin{array}{l}Here\,i=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\end{array} \)
and sixth root of z
i.e. zr = z1/6 .
\(\begin{array}{l}z=64\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\end{array} \)
,
\(\begin{array}{l}{{z}_{r}}={{z}^{1/6}}\end{array} \)
,
\(\begin{array}{l}=2\left( \cos \frac{2r\pi +\frac{\pi }{2}}{6}+\sin \frac{2r\pi +\frac{\pi }{2}}{6} \right)\end{array} \)
Where r = 0, 1, 2, 3, 4, 5
The roots z0 , z1 , z3 , z4 , z5 are given by
\(\begin{array}{l}{{z}_{0}}=2\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right)\end{array} \)
,
\(\begin{array}{l}{{z}_{1}}=2\left( \cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12} \right)\end{array} \)
,
\(\begin{array}{l}{{z}_{2}}=2\left( \cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12} \right)\end{array} \)
,
\(\begin{array}{l}{{z}_{3}}=2\left( \cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12} \right)=-2\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right)\end{array} \)
,
\(\begin{array}{l}{{z}_{4}}=2\left( \cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12} \right)=-2\left( \cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12} \right)\end{array} \)
,
\(\begin{array}{l}{{z}_{5}}=2\left( \cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12} \right)=-2\left( \cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12} \right)\end{array} \)
Illustration 21: 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
\(\begin{array}{l}\frac{{{z}_{2}}-{{z}_{3}}}{{{z}_{1}}-{{z}_{3}}}={{e}^{-i\pi /2}}.\end{array} \)
Therefore solve it using algebra method.
Given CB = CA and angle
\(\begin{array}{l}\angle C=\frac{\pi }{2}\end{array} \)
,
\(\begin{array}{l}\frac{{{z}_{2}}-{{z}_{3}}}{{{z}_{1}}-{{z}_{3}}}={{e}^{-i\pi /2}}\end{array} \)
or (z3 – z2 )2 = i2 (z1 – z3 )2
(z3 – z2 )2 = -(z1 – z3 )2
Or
\(\begin{array}{l}z_{3}^{2}+z_{2}^{2}-2{{z}_{2}}{{z}_{3}}+z_{1}^{2}+z_{3}^{2}-2{{z}_{1}}{{z}_{3}}=0\end{array} \)
Add and subtract 2z1 z2 , we get
\(\begin{array}{l}z_{1}^{2}+z_{2}^{2}-2{{z}_{1}}{{z}_{2}}+2z_{3}^{2}+2{{z}_{2}}{{z}_{3}}-2{{z}_{1}}{{z}_{3}}+2{{z}_{1}}{{z}_{2}}=0,\,or\end{array} \)
\(\begin{array}{l}{{({{z}_{1}}-{{z}_{2}})}^{2}}+2({{z}_{3}}-{{z}_{1}})({{z}_{3}}-{{z}_{2}})=0,\,or\,{{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})({{z}_{3}}-{{z}_{2}}).\end{array} \)
Illustration 22: If A, B, C be the angles of triangle then
\(\begin{array}{l}prove\,that\,\left| \begin{matrix} {{e}^{2iA}} & {{e}^{-iC}} & {{e}^{-iB}} \\ {{e}^{-iC}} & {{e}^{2iB}} & {{e}^{-iA}} \\ {{e}^{-iB}} & {{e}^{-iA}} & {{e}^{2iC}} \\ \end{matrix} \right|is\,purely\,real.\end{array} \)
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)
\(\begin{array}{l}\,\left| \begin{matrix} {{e}^{iA}} & {{e}^{-i(A+C)}} & {{e}^{-i(A+B)}} \\ {{e}^{-i(B+C)}} & {{e}^{iB}} & {{e}^{-i(B+A)}} \\ {{e}^{-i(B+C)}} & {{e}^{-i(C+A)}} & {{e}^{iC}} \\ \end{matrix} \right|=-1\left| \begin{matrix} {{e}^{iA}} & -{{e}^{iB}} & -{{e}^{i}} \\ -{{e}^{iA}} & {{e}^{iB}} & -{{e}^{iC}} \\ -{{e}^{iA}} & -{{e}^{iB}} & {{e}^{iC}} \\ \end{matrix} \right|,by\,(ii)\end{array} \)
Take eiA , eiB and eiC common from C1 , C2 and C3 and again put ei(A+B+C) =eiπ = –1
∴
\(\begin{array}{l}\,\,\,\,\Delta =(-1)(-1)\left| \begin{matrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right|\end{array} \)
Now make two zeros and expand Δ = –4 which is purely real.
Illustration 23 : Show that all the roots of the equation
\(\begin{array}{l}{{z}^{n}}cos{{q}_{0}}+{{z}^{n-1}}cos\text{ }{{q}_{1}}+{{z}^{n-2}}cos\,{{q}_{2}}+\ldots ..+z\,cos\text{ }{{q}_{n-1}}+cos\text{ }{{q}_{n}}=2\end{array} \)
lie outside the circle
\(\begin{array}{l}|z|=\frac{1}{2}\end{array} \)
where q0 , q1 etc are real.
Sol: By using triangle inequality.
Here
\(\begin{array}{l}\left| {{z}^{n}}cos\text{ }{{q}_{0}}+{{z}^{n-1}}cos\text{ }{{q}_{1}}+{{z}^{n-2}}\text{ }cos\text{ }{{q}_{2}}\text{ }+\text{ }\ldots .\text{ }+\text{ }z\text{ }cos\text{ }{{q}_{n-1}}+cos\text{ }{{q}_{n}} \right|=2\end{array} \)
By triangle inequality.
2=
\(\begin{array}{l}|{{z}^{n}}cos{{\theta }_{n}}+{{2}^{n-1}}cos\,{{q}_{1}}+{{2}^{n-1}}\,\cos \,{{q}_{2}}+…+z\cos \,{{q}_{n-1}}+\cos \,{{q}_{n}}|\le |{{z}^{n}}cos{{q}_{n}}|+|{{z}^{n-1}}cos\,{{q}_{1}}|\end{array} \)
+
\(\begin{array}{l}|{{z}^{n-2}}cos\,{{q}_{2}}|+….++|zcos\,{{q}_{n-1}}|+|cos\,{{q}_{n}}|=|{{z}_{n}}||cos\,{{q}_{n}}|+|{{z}^{n-1}}||cos\,{{q}_{n}}|+…+|z|cos\,{{q}_{n-1}}|+\end{array} \)
\(\begin{array}{l}|cos\,{{q}_{n}}|\,\le \,|z{{|}^{n}}+|z{{|}^{n-1}}+….+|z|+1\end{array} \)
\(\begin{array}{l}\left( Since,\;\,|cos\,{{q}_{1}}|\le 1\,and\,|{{z}^{n-1}}|=|z{{|}^{n+1}} \right)\end{array} \)
\(\begin{array}{l}=\frac{1-|z{{|}^{n+1}}}{1-|z|}<\frac{1}{1-|z|}\end{array} \)
∴
\(\begin{array}{l}2<\frac{1}{1-|z|}\end{array} \)
\(\begin{array}{l}sol\,1-|z|is\,positive\,and\,1-|z|\,<\frac{1}{2}\end{array} \)
∴
\(\begin{array}{l}\,|z|>1-\frac{1}{2}=\frac{1}{2}\end{array} \)
∴
\(\begin{array}{l}\,\,\,All\,z\,satisfying\,(i)\,lie\,outside\,the\,circle\,|z|=\frac{1}{2}\end{array} \)
Illustration 24: If
\(\begin{array}{l}z+\frac{1}{z}=2\cos \theta ,prove\,that\,\left| \frac{{{z}^{2n}}-1}{{{z}^{2n}}+1} \right|=|tan\,nq|\end{array} \)
Sol: By using formula of roots of quadratic equation, we can solve this problem.
Here z + 1/z = 2 cos θ
∴
\(\begin{array}{l}\,{{z}^{2}}-2\cos \,\theta .\,z+1=0\end{array} \)
∴
\(\begin{array}{l}\,z=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}=\cos \theta \pm i\sin \theta\end{array} \)
Taking positive sign,
∴
\(\begin{array}{l}\,\frac{1}{z}={{(cos\,\theta \,+\,i\,sin\,\theta )}^{-1}}=cos\,\theta \,-\,i\,sin\,\theta\end{array} \)
∴
\(\begin{array}{l}\,\frac{{{z}^{2n}}-1}{{{z}^{2n}}+1}=\frac{{{z}^{n}}-\frac{1}{{{z}^{n}}}}{{{z}^{n}}+\frac{1}{{{z}^{n}}}}=\frac{{{(cos\theta +isin\theta )}^{n}}-{{(cos\theta -isin\theta )}^{n}}}{{{(cos\theta +isin\theta )}^{n}}+{{(cos\theta -isin\theta )}^{n}}}\end{array} \)
\(\begin{array}{l}=\frac{\cos n\theta +i\sin \theta -(cosn\theta -isin\theta )}{\cos n\theta +i\sin \theta +(cosn\theta -isinn\theta }=\frac{2i\,\sin n\theta }{2\cos n\theta }=i\tan n\theta\end{array} \)
.
Taking negative sign,
Similarly, we get
\(\begin{array}{l}\frac{{{z}^{2n}}-1}{{{z}^{2n}}+1}=\frac{-2i\sin \,n\theta }{2\cos n\theta }=-i\tan \,n\theta\end{array} \)
∴
\(\begin{array}{l}\,\left| \frac{{{z}^{2n}}-1}{{{z}^{2n}}+1} \right|=|\pm i\,tan\,nq|=|tan\,nq|,\end{array} \)
Illustration 25: 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 θ + i sin θ, therefore by obtaining modulus of z we can solve above problem.
|z – 2 + 2i| = 1
\(\begin{array}{l}\Rightarrow \,z-2+2i=\cos \,\theta +i\,\sin \,\theta\end{array} \)
Where θ is some real numbers.
\(\begin{array}{l}\Rightarrow \,z=(2+cos\theta )+(sin\theta -2)i\end{array} \)
\(\begin{array}{l}\Rightarrow \,|z|={{[{{(2+cos\,\theta )}^{2}}+{{(sin\theta -2)}^{2}}]}^{1/2}}\end{array} \)
\(\begin{array}{l}={{[8+co{{s}^{2}}\theta +si{{n}^{2}}\theta +4(cos\theta -\sin \theta )]}^{1/2}}\end{array} \)
\(\begin{array}{l}=\,{{\left[ 9+4\sqrt{2}\cos \left( \theta +\frac{\pi }{4} \right) \right]}^{1/2}}\end{array} \)
|z| will be least if cos (θ + π/4) is least, that is, if cos (θ + π/4) = -1 or θ =
\(\begin{array}{l}\frac{3\pi }{4}\end{array} \)
. Thus, least value of |z| is
\(\begin{array}{l}{{\left( 9-4\sqrt{2} \right)}^{1/2}}for\,z=\left( 2-\frac{1}{\sqrt{2}} \right)+i\left( \frac{1}{\sqrt{2}}-2 \right)\end{array} \)
Illustration 26: The area of the triangle whose vertices are represented by the complex numbers 0, z,
\(\begin{array}{l}z{e}^{i\alpha},(0<\alpha <\pi) \text \ equals\\\end{array} \)
Solution:
Vertices are
\(\begin{array}{l}0=0+i0,z=x+iy \text \ and \ z{{e}^{i\alpha}}\\=(x+iy)\,\,(\cos \alpha +i\sin \alpha)=(x\cos \alpha -y\sin \alpha )+i(y\cos \alpha +x\sin \alpha )\\ Area =\frac{1}{2}\,\left| \,\begin{matrix} 0 & 0 & 1 \\ x & y & 1 \\ (x\cos \alpha -y\sin \alpha )\,\,\, & (y\cos \alpha +x\sin \alpha )\,\, & 1 \\ \end{matrix}\, \right|\\=\frac{1}{2}[xy\cos \alpha +{{x}^{2}}\sin \alpha -xy\cos \alpha +{{y}^{2}}\sin \alpha ]\\=\frac{1}{2}\sin \alpha ({{x}^{2}}+{{y}^{2}})\\=\frac{1}{2}|z{{|}^{2}}\sin \alpha [\,\,|z|=\sqrt{{{x}^{2}}+{{y}^{2}}}].\\\end{array} \)
Illustration 27: The complex numbers z = x + iy which satisfy the equation
\(\begin{array}{l}\left| \frac{z-5i}{z+5i} \right|=1 \text \ lie \ on \\\end{array} \)
Solution:
\(\begin{array}{l}\left| \frac{z-5i}{z+5i} \right|=1\\ \left| \frac{x+i(y-5)}{x+i(y+5)} \right|=1 \\ |x+i(y-5)|\,=\,|x+i(y+5)|, \left(\left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=\frac{|{{z}_{1}}|}{|{{z}_{2}}|} \right) \\ {{x}^{2}}+25-10y+{{y}^{2}}={{y}^{2}}+{{x}^{2}}+25+10y \\ 20y=0 \\ y=0\\\end{array} \)
Illustration 28: The region of the complex plane for which
\(\begin{array}{l}\left| \frac{z-a}{z+\overline{a}} \right|=1\,\,[R(a)\ne 0] \text \ is\\\end{array} \)
Solution:
We have
\(\begin{array}{l}\left| \frac{z-a}{z+\bar{a}} \right|=1 \\ |z-a|\,=\,|z+\overline{a}| \\ |z-a{{|}^{2}}=|z+\overline{a}{{|}^{2}} \\ (z-a)(\overline{z-a})=(z+\overline{a})(\overline{z+\overline{a}}) \\ (z-a)(\overline{z}-\overline{a})=(z+\overline{a})(\overline{z}+a) \\ z\overline{z}-z\overline{a}-a\overline{z}+a\overline{a}=z\overline{z}+za+\overline{a}\overline{z}+\overline{a}a \\ za+z\overline{a}+\overline{a}\overline{z}+a\overline{z}=0\, \Rightarrow (a+\overline{a})(z+\overline{z})=0 \\ z+\overline{z}=0\,\,(a+\overline{a}=2 {Re}(a)\ne 0) \\ 2 {Re}(z)=0\\ 2x=0 \\ x=0\end{array} \)
which is the equation of y-axis.
Illustration 29:
\(\begin{array}{l}\frac{3+2 i \sin \theta}{1-2 i \sin \theta} \text \ will \ be \ purely \ imaginary, \ if \ \theta equals\\ (1) 2 \mathrm{n} \pi \pm \frac{\pi}{3}\\ (2) \mathrm{n} \pi+\frac{\pi}{3}\\ (3) \mathrm{n} \pi \pm \frac{\pi}{3}\\ (4) \text { None of these}\\ Solution: \frac{3+2 i \sin \theta}{1-2 i \sin \theta} \text \ will \ be \ purely \ imaginary, \ if \ the \ real \ part \ vanishes, \\ \begin{array}{l} \frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=0 \\ \Rightarrow 3-4 \sin ^{2} \theta=0 \quad \text { (only if } \theta \text { be real) } \\ \Rightarrow \sin \theta=\pm \frac{\sqrt{3}}{2}=\sin \left(\pm \frac{\pi}{3}\right) \\ \Rightarrow \theta=n \pi+(-1)^{n}\left(\pm \frac{\pi}{3}\right)=n \pi \pm \frac{\pi}{3} \end{array}\\ Answer: [3]\end{array} \)
Illustration 30:
\(\begin{array}{l}\text \ If \ \frac{\tan \theta-i\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)}{1+2 i \sin \frac{\theta}{2}} \ is \ purely \ imaginary \ then \ general \ value \ of \ \theta \ is -\\ (1) \mathrm{n} \pi+\frac{\pi}{4}\\ (2) 2 n \pi+\frac{\pi}{4}\\ (3) \mathrm{n} \pi+\frac{\pi}{2}\\ (4) 2 n \pi+\frac{\pi}{2}\\ Solution: \text { Multiply above and below by conjugate of denominator and put real part equal to zero.}\\ \begin{array}{l} =\frac{\tan \theta-i\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)}{1+2 i \sin \frac{\theta}{2}} \times \frac{1-2 i \sin \frac{\theta}{2}}{1-2 i \sin \frac{\theta}{2}} \\ =\frac{\tan \theta-2 \sin \frac{\theta}{2}\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)-i\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}+2 \tan \theta \sin \frac{\theta}{2}\right)}{1+4 \sin ^{2} \frac{\theta}{2}} \\ \therefore \tan \theta-2 \sin \frac{\theta}{2}\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)=0 \\ \Rightarrow \frac{\sin \theta}{\cos \theta}-(1-\cos \theta)-\sin \theta=0 \\ \Rightarrow \sin \theta\left(\frac{1-\cos \theta}{\cos \theta}\right)-(1-\cos \theta)=0 \\ \Rightarrow(1-\cos \theta)(\tan \theta-1)=0 \\ \cos \theta=1 \Rightarrow \theta=2 n \pi \text { and } \tan \theta=1 \Rightarrow \theta=n \pi+\frac{\pi}{4} \end{array}\\ Answer: [1]\end{array} \)
Illustration 31:
\(\begin{array}{l}\sum_{\mathrm{k}=1}^{6}\left(\sin \frac{2 \pi \mathrm{k}}{7}-\mathrm{i} \cos \frac{2 \pi \mathrm{k}}{7}\right) \text { is equal to}\\ (1) -1\\ (2) 0\\ (3) i\\ (4) -i\\ Solution:\\ \text { Expression } =-i \sum_{k=1}^{6}\left(\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}\right) \\ =-i \sum_{k=1}^{6} e^{\frac{2 \pi k}{7} i} \\ =-i\left[e^{\frac{2 \pi}{7} i}+e^{\frac{4 \pi}{7} i}+\ldots \ldots e^{\frac{12 \pi}{7}}\right] \\ =-i e^{\frac{2 \pi}{7}} \frac{\left(1-e^{\frac{12 \pi}{7} i}\right)}{\left(1-e^{\frac{2 \pi}{7} i}\right)}=-i \frac{e^{\frac{2 \pi}{7}}-e^{2 \pi i}}{1-e^{\frac{2 \pi}{7}}}=i\left(\frac{1-e^{\frac{2 \pi}{7}}}{1-e^{\frac{2 \pi}{7}}}\right) \\ =i\\ Answer: [3]\end{array} \)
Quadratic Equations and Complex Numbers Important JEE Main Questions Also Check:
Geometry of Complex Numbers
Complex Numbers Class 11
Complex Numbers – Important Topics
Complex Numbers – Important Questions
