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, students will learn about complex numbers – definition, standard form, algebraic operations, conjugate, complex numbers’ polar form, Euler’s form and much more. A complex number is a combination of a real number and an imaginary number. 

Table of Contents

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

What Are Complex Numbers?

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

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

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

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

Integral Powers of an Iota

An imaginary number, i (iota), is defined as √-1 since i = √-1; we have i² = –1, i³ = –i, and i⁴ = 1.

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

Let q be the quotient and r be the remainder.

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

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

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

iⁿ + iⁿ⁺¹ + i^{n + 2} + i^{n + 3} = 0, n ∈ z

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

Illustration 1: Evaluate i²⁰¹

Solution: 201 leaves the 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:

= (2i/2)⁴ⁿ + (-2i/2)⁴ⁿ = i⁴ⁿ + (-i)⁴ⁿ

= 1 + 1 = 2

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

Solution:

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

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 the quotient of a + bi and c + di is given by

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

Conjugate of a Complex Number

Let = z = a + ib be a complex number. We define the 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 a Complex Number

1. \(\begin{array}{l}{{z}_{1}}={{z}_{2}}\Leftrightarrow {{\overline{z}}_{1}}={{\overline{z}}_{2}}[/latex
2. \(\begin{array}{l}\overline{(\bar{z})}=z\end{array} \)
3. \(\begin{array}{l}z+\overline{z}=2\,{Re}(z)\end{array} \)
4. \(\begin{array}{l}z-\overline{z}=2i\,{Im}\,(z)\end{array} \)
5. \(\begin{array}{l}z=\overline{z}\Leftrightarrow z\ \text{is purely real}\end{array} \)
6. \(\begin{array}{l}z+\overline{z}=0\Leftrightarrow z\ \text{is purely imaginary.}\end{array} \)
7. \(\begin{array}{l}z\overline{z}={{[Re\,(z)]}^{2}}+{{[Im(z)]}^{2}}\end{array} \)
8. \(\begin{array}{l}\overline{{{z}_{1}}+{{z}_{2}}}={{\overline{z}}_{1}}+{{\overline{z}}_{2}}\end{array} \)
9. \(\begin{array}{l}\overline{{{z}_{1}}-{{z}_{2}}}={{\overline{z}}_{1}}-{{\overline{z}}_{2}}\end{array} \)
10. \(\begin{array}{l}\overline{{{z}_{1}}{{z}_{2}}}={{\overline{z}}_{_{1}}}{{\overline{z}}_{_{2}}}\end{array} \)
11. \(\begin{array}{l}\overline{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}=\frac{{{\overline{z}}_{1}}}{{{z}_{2}}}\ \text{if}\ z_2 \ne 0\end{array} \)
12. If P(z) = a₀ + a₁ z + a₂ z² + …. + a_n zⁿ
    Where a₀, a₁, ….. a_n and z are complex number, then
    \(\begin{array}{l}\overline{P(z)}={{\overline{a}}_{0}}+{{\overline{a}}_{1}}(\overline{z})+{{\overline{a}}_{2}}{{(\overline{z})}^{2}}+….+{{\overline{a}}_{n}}{{(\overline{z})}^{n}} = \overline{P}(\overline{z})\end{array} \)
    Where
    \(\begin{array}{l}\overline{P}(z)={{\overline{a}}_{0}}+{{\overline{a}}_{1}}z+{{\overline{a}}_{2}}{{z}^{2}}+….+{{\overline{a}}_{n}}{{z}^{n}}\end{array} \)
13. \(\begin{array}{l}\text{If}\ R(z) = \frac{P(z)}{Q(z)}\end{array} \)
    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 and z₂ are complex numbers, then |az₁ + bz₂ |² + |bz₁ – az₂ |² = (a² + b²) (|z₁|² + |z₂|²)

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

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

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

(xiv) |z₁ – z₂| ≤ |z₁| + |z₂|

(xv) ||z₁| – |z₂|| ≤ |z₁| + |z₂|

(xvi) |z₁ – z₂| ≥ ||z₁| – |z₂||

Square Root of a Complex Number

Let z = x + iy then

Where

NOTE:

(i)

(ii)

(iii)

Modulus and Argument of a Complex Number

Let z = x + iy = (x, y) for all x, y ∈ R and

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

i.e., OP = r = |z|

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

or arg (z) =

Also, the 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.

The argument of z will be θ, π – θ, π + θ and 2π – θ according to the point z lies in I, II, III and IV quadrants, respectively, where

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

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

And

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

(i) (x, y) ∈ first quadrant x > 0, y > 0.

The principal value of arg (z) = θ

It is an acute angle and positive.

(ii) (x, y) ∈ second quadrant x < 0, y > 0.

The principal value of arg (z) = θ

(iii) (x, y) ∈ third quadrant x < 0, y < 0.

The principal value of arg (z) = θ

It is an obtuse angle and negative.

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

The principal value of arg (z) = θ

It is an acute angle and negative.

Polar Form of a Complex Number

We have, z = x + iy

= |z| [cos θ + i sin θ]

Where |z| is the modulus of the complex number, i.e., the distance of z from the origin, and θ is the argument or amplitude of the complex number.

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

Euler’s Form of a Complex Number

e^iθ = cos θ + i sin θ

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

z = x + iy (Cartesian form)

= |z| [cos θ + I sin θ] (polar form)

= |z| e^iθ

De Moivre’s Theorem and Its Applications

(a) De Moivre’s theorem for integral index. If n is a integer, then (cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ)

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

(cos θ + i sin θ)ⁿ is cos (nθ) + i sin (nθ).

In fact, if n = p/q where p, q ∈ I, q > 0 and p,q have no factors in common, then (cos θ + i sin θ)ⁿ has q distinct values, one of which is cos (nθ) + i sin (nθ)

The nth Roots of Unity

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

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

Where k ϵ I and

Where, k = 0, 1, 2, …., n -1.

Notation

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

The sum of the roots of unity is zero.

We have

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

Also, note that

nth Root of Unity Video Lesson

Cube roots of unity

Cube roots of unity are given by 1, ω, ω², where

Some Results Involving Complex Cube Root of Unity (ω)

(i) ω³ = 1

(ii) 1 + ω + ω² = 0

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

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

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

(vi) a² + b² + c² – bc – ca – ab

= (a + bω + cω²) (a + bω² + cω)

(vii) a³ + b³ + c³ – 3abc

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

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

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

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

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

Solutions of y³ = 1 are 1, ω, ω².

x = a^1/3, a^1/3 ω, a^1/3 ω².

Cube Roots of Unity Video Lesson

Logarithm of a Complex Number

Log_e(x + iy) = log_e (|z|e^iθ)

= loge |z| + log_e e^iθ

= loge |z| + iθ

Problems on Complex Numbers

Illustration 1: The number of solutions of

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

Ans. (d)

Solution

⇒ |z| = 0 or |z| = 1 [Since, |z| + 1 > 0]

If |z| = 0, then z = 0

Thus,

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 the value of the expression

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

(d) none of these

Ans. (a)

Solution: The r^th term of the given expression is

(r + 1) (r + ω) (r + ω²) = r³ + 1

The value of the given expression is

Illustration 3: Find the real part of (1 – i)^-i.

Sol: Let z = (1 – i)^-i

Taking log on both sides, we have

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

Sol. We have, x³ – 3x² + 3x + 7 = 0

Then,

Illustration 5: If z₁ and z₂ are 1 – i, -2 + 4i, respectively.

Sol.

Illustration 6: Find the square root of z = -7 – 24i.

Sol. Consider z₀ = x + iy be a square root then;

or

Equating real and imaginary parts, we get

and 2xy = -24

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

Illustration 7: If n is a positive integer and ω be an imaginary cube root of unity, prove that

Sol. Case: I.

Case: II.

(a) Let n = 3m + 1

(b) Let n = 3m + 2

Illustration 8: Show that

Sol. Consider z = x + iy

or

which represents a circle.

Illustration 9: If |z₁| = |z₂| = …. = |z_n| = 1, prove that

Sol.

L.H.S.

Illustration 10: If |z₁ + z₂| = |z₁ – z₂|, prove that arg(z₁) – arg(z₂) = odd multiple of π/2.

Sol. As we know

or

Let

then

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 + a₁x + a₂x²+ ….+ a_nxⁿ, then show that

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

z = r (cos θ + i sin θ) and (cos θ + i sin θ)ⁿ

= cos nθ + i sin nθ, we can solve this problem.

Put x = i in the given expansion

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

Equating real and imaginary parts,

Therefore, (a) and (b) are proved.

Illustration 13: Solve the equation

Sol: Apply modulus on both sides.

Illustration 14: If z = x + iy and

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

Sol: Substitute value of ω in |ω| = 1

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

Hence, z lies on the real axis.

Illustration 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 the greatest and least values of |z + 1| are,

(b) 6, 1

(d) 3, 1

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

Illustration 16: Find the equation of the circle for which arg

Sol:

represent a major arc of the circle, of which the line joining (6, 2)) and (2, 2) is a chord that subtends an angle π/4 at the circumference.

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

The equation of the required circle is

Illustration 17: if |z| > 3, prove that the least value of

Sol: Here,

Now, |z| > 3

Adding the two like inequalities

Hence, from (i) and (ii), we get

∴ The least value is 8/3.

Illustration 18: If z₁, z₂, z₃ are non-zero complex numbers such that

|z₁| = |z₂| = |z₃|.

Sol: Use algebra to solve this problem.

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

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

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

This implies that the centroid is also at the origin, so 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

Equating real and imaginary parts, we get

y + 1 = 0 y = –1 ….(i)

and

By solving (i) and (ii), we get

Or (c² – 1)x² + 2c²x + 2c² = 0

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

As x is real and c > 1, we have:

(Thus, for √2, 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:

i.e., z_r = z^1/6.

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

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

Illustration 21: Let three vertices, A, B, and C (taken in clockwise order) of an isosceles right-angled triangle with a right angle at C, be affixes of complex numbers z₁, z₂, z_3, respectively. Show that (z₁ – z₂)² = 2(z₁ – z₃) (z₃ – z₂).

Sol: Here,

Therefore, solve it using the algebra method.

Given CB = CA and angle ∠C = π/2,

(z₃ – z₂)² = -(z₁ – z₃)²

Or

Add and subtract 2z₁z₂, and we get

Illustration 22: If A, B, C be the angles of a triangle, then

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

e^-pi = –1 …. (i)

e^i(B+C) = e^i(π–A) = e^pi e^–iA = –e^–iA

e^i(B+C)= –e^iA …. (ii)

Take e^iA, e^iB and e^iC common from R₁, R₂ and R_3, respectively. Δ = e^i(A+B+c)

Take e^iA, e^iB and e^iC common from C₁, C₂ and C₃ and again put e^i(A+B+C)=e^iπ = –1

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

Illustration 23: Show that all the roots of the equation

lie outside the circle |z| = 1/2 where q₀, q₁ etc. are real.

Sol: By using triangle inequality,

By the triangle inequality,

∴ All z satisfying (i) lie outside the circle, |z| = 1/2.

Illustration 24: If z + (1/z) = 2 cos θ, prove that

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

Here, z + 1/z = 2 cos θ

Taking positive the sign,

Taking the negative sign,

Similarly, we get

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 the modulus of z, we can solve the above problem.

|z – 2 + 2i| = 1

Where θ is some real numbers.

|z| will be least if cos (θ + π/4) is least, that is, if cos (θ + π/4) = -1 or θ = 3π/4.

Thus, the least value of |z| is

Illustration 26: The area of the triangle whose vertices are represented by the complex numbers 0, z,

Sol:

Vertices are

Illustration 27: The complex numbers z = x + iy which satisfy the equation

Sol:

Illustration 28: The region of the complex plane for which

Sol:

We have

2x = 0

x = 0, which is the equation of y-axis.

Illustration 29:

Sol:

Illustration 30:

Sol:

Multiply above and below by the conjugate of the denominator and put the real part equal to zero.

Illustration 31:

Sol:

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