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.
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 ofRe(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
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Illustration 1: Evaluate i201
Solution: 201 leaves remainder as 1 when it is divided by 4 therefore i201 = i1 = i
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=z2⇔z1=z2
(ii)(zˉ)=z
(iii)z+z=2Re(z)
(iv)z−z=2iIm(z)
(v)z=z⇔z is purely real
(vi)z+z=0⇔z is purely imaginary.
(vii)zz=[Re(z)]2+[Im(z)]2
(viii)z1+z2=z1+z2
(ix)z1−z2=z1−z2
(x)z1z2=z1z2
(xi)(z2z1)=z2z1 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
= P(z)
Where P(z)=a0+a1z+a2z2+….+anzn
(xiii) If R(z) = Q(z)P(z) where P (z) and Q (z) are polynomials in z, and Q(z) ≠ 0, then
R(z)=Q(z)P(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|.
Let z = x + iy = (x, y) for all x, y∈R and i = −1
The length OP is called modulus of the complex number z denoted by |z|,
i.e. OP = r = |z| = (x2+y2)
and if (x, y) ≠ (0, 0), then θ is called the argument or amplitude of z,
i.e. θ = tan−1(xy) [angle made by OP with positive X-axis]
or arg (z) = tan−1(y/x)
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 ∈ 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∣∣∣xy∣∣∣.
Illustration. Find the arguments of z1 = 5 + 5i, z2 = –4 + 4i, z3 = –3 – 3i and z4 = 2 – 2i, where
i=−1.
Solution: Since, z1, z2, z3 and z4 lies in I, II, III and IV quadrants respectively. The arguments are given by
And arg (z4 ) = 2π−tan−1∣∣∣2−2∣∣∣=2π+tan−11=2π−4π=47π
Principal value of the Argument
The value θ of the argument which satisfies the inequality −π<θ≤π is called the principal value of the argument.
If x = x + iy = ( x, y), ∀ x, y ∈ R and i=−1, then
Arg(z) = tan−1(xy) always gives the principal value. It depends on the quadrant in which the point (x, y) lies.
(i) (x, y) ∈ first quadrant x > 0, y > 0.
The principal value of arg (z) = θ=tan−1(xy)
It is an acute angle and positive.
(ii) (x, y) ∈ second quadrant x < 0, y > 0.
The principal value of arg (z) = θ = π−tan−1(∣x∣y). It is an obtuse angle and positive.
(iii) (x, y) ∈ third quadrant x < 0, y < 0.
The principal value of arg (z) = θ = −π+tan−1(xy)
It is an obtuse angle and negative.
(iv) (x, y) ∈ fourth quadrant x > 0, y < 0.
The principal value of arg (z) = θ = −tan−1(x∣y∣)
It is an acute angle and negative.
Polar Form of a Complex Number
We have, z = x + iy
=x2+y2[x2+y22+ix2+y2x]
= |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 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[qp(2kπ+θ)]+isin[qp(2kπ+θ)]
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(n2kπ)+isin(n2kπ) [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(n2π)+isin(n2π)
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 = 1−ω1−ωn
But ωn = 1 as ω is a nth root of unity.
∴ 1+ω+…+ωn−1=0
Also, note that
x−11+x−ω1….+x−ωn−11=xn−1nxn−1
Cube roots of Unity
Cube roots of unity are given by 1, ω, ω2, where ω=cos(32π)+isin(32π)=2−1+3iandω2=2−1−3i
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|eiƟ)
= loge |z| + loge eiƟ
= loge |z| + iƟ
= loge(x2+y2)+iarg(z)
∴ loge(z)=loge∣z∣+iarg(z)
Problems on Complex Numbers
Illustration: The number of solutions of z3+z=0 is
Illustration: If ∣z1+z2∣=∣z1−z2∣, prove that argz1−argz2= odd multiple of 2π.
Sol. As we know ∣z∣=z.zˉ. Apply this formula and consider z=r(cosθ+isinθ).∣z1+z2∣2=∣z1−z2∣2⇒(z1+z2)(zˉ1+zˉ2)=(z1−z2)(zˉ1−zˉ2)orz1zˉ1+z2zˉ2+z2zˉ1+z1zˉ2=z1zˉ1+z2zˉ2−z2zˉ1−z1zˉ2
or 2(z2zˉ1+z1zˉ2)=0;Re(z1zˉ2)=0
Let z1=r1(cosθ1+isinθ1)andz2=r2(cosθ2+isinθ2);
then z1zˉ2=r1r2(cos(θ1−θ2)+isin(θ1−θ2))
∴ cos(θ1−θ2)=0(asRe(z1zˉ2)=0)θ1−θ2= odd multiple of 2π.
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) a0−a2+a4+….=22ncos4nπ
(b) a1−a3+a5+….=22nsin4nπ
Sol: simply put x = i in the given expansion and then by using formula
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+17,17−3
(b) 6, 1
(c) 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.
∴ Greatestdistance=3+12+42=3+17
Least distance = 12+42−3=17−3
Illustration: Find the equation of the circle for which arg (z−2−2iz−6−2i)=π/4.
Sol:arg(z−2−2iz−6−2i)=π/4
represent a major arc of circle of which Line joining (6, 2)) and (2, 2) is a chord that subtends an angle 4π at circumference.
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 Equation of required circle is
∣z−4−4i∣=22
Illustration: if |z| > 3, prove that the least value of ∣∣∣z+z1∣∣∣is38