Complex Number Formulas: Properties & De Moivre’s Theorem

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A complex number is a number with both real and imaginary parts that can be expressed in the form a + bi, where a and b are real numbers and i is an imaginary unit, satisfying the equation i² = −1. In this expression, b is the imaginary part and a is the real part of the complex number.

Complex numbers are formed by combining real numbers and imaginary numbers.
A complex number is a two-dimensional complex plane that expands the notion of a one-dimensional number line by employing a vertical axis for the imaginary component and a horizontal axis for the real part.
The concept of a two-dimensional complex plane is explained in detail with the help of complex numbers using the horizontal axis for the real part and the vertical axis for the imaginary part.

Read More: Algebra Formulas for Class 10

Table of Content

What is a Complex Number?
Equality of Complex Number Formula
Powers of Complex Numbers
Terms used in Complex Numbers
Conjugate of Complex Number
Properties of Conjugate
Modulus of a Complex Number
De Moivre’s Theorem
Proof of De Moivre’s Theorem
Solved Examples
Things to Remember
Sample Questions

Key Terms: Complex Number, De Moivre’s Theorem, Powers of Complex Numbers.

What is a Complex Number?

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A complex number is a number with both real and imaginary parts that can be expressed in the form x + iy, where x and y are real numbers and i is the imaginary part.

A complex number extends the concept of a one-dimensional number line to the two-dimensional complex plane, using a vertical axis for the imaginary part and a horizontal axis for the real part.
The square root of a negative real number is known as an imaginary number, for example, √-2, √-5, etc.
The quantity √-1 is an imaginary unit and it is represented by the symbol ‘i’.
They may be used to simulate periodic movements and also in regulating alternating currents.

The video below explains this:

Complex Numbers Detailed Video Explanation:

Equality of Complex Number Formula

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Considering the equation a+bi = c+di, in which real part is equal with each other and imaginary parts are equal i.e., a=c and b=d.

Addition of Complex Numbers: (a + bi) +(c + di) = (a + c) + (b + d) i.
Subtraction of Complex Numbers: (a + bi) − (c + di) = (a − c) + (b − d) i.
Multiplication of Complex Numbers: (a + bi) × (c + di) = (ac − bd) + (ad + bc) i.
Multiplication Conjugates: (a + bi) × (a + bi) = a²+ b².
Division of Complex Numbers: (a + ib) / (c + id) = (ac + bd) / (c² + d²) + i (bc – ad) / (c² + d²).

Read More: Binomial Theorem

Powers of Complex Numbers

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The powers of complex numbers are mentioned below:

i = √-1

i² = – 1

i³ = i . i² = i (-1) = – i

i⁴ = (i²)² = (-1)² = 1

i⁴ⁿ = 1

i^{4n + 1} = i

i^{4n + 2} = – 1

i^{4n + 3} = – i

Read More: Radians.

Terms Used in Complex Numbers

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The following are the few terms that are used in complex numbers:

Argument is the angle formed between the positive real axis and the segment connecting the origin to the complex number plot in the complex plane.
For a given complex number a + bi, the complex conjugate is a – bi.
It is a plane having two perpendicular axes, on which the complex number a + bi is plotted with coordinates as (a, b).
In the complex plane, the modulus is the distance between the complex number plot and the root.
The axis in the complex plane generally coincides with the y-axis of the rectangular coordinate system and on which the imaginary part bi of the complex number a + bi is plotted.
Polar form of a complex number - Let z be the complex number a + bi,
Polar form of z = r(cos(Θ)+isin(Θ)), where r = | z| and Θ is the argument of z.
The real axis in the complex plane is the axis that corresponds with the x-axis of the rectangular coordinate system. The real component of the complex number a + bi is shown on the real axis.

Also Read:

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Conjugate of Complex Number

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Conjugate of a complex number consist of one part with the original real number but the other part with the opposite sign of the imaginary number.

It is found just like the conjugate of any other number.
The conjugate of a complex number a is represented as ā.
Let z = x + iy, if ‘i’ is replaced by (-i), then said to be conjugate of the complex number z and it is denoted by Z, i.e., Z = x – iy.

Read More: Value of e

Properties of Conjugate

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The following are the properties of conjugate:

(\(\bar{\bar{z}}\)) = z
z₁ \(\bar{+}\) z₂ = \(\bar{z_1} + \bar{z_2}\)
\(\bar{z_1 - z_2} = \bar{z_1} - \bar{z_2}\)
\(\bar{z_1z_2} = \bar{z_1}\bar{z_2}\)
\((\bar{\frac{z_1}{z_2}}) = \frac{\bar{z_1}}{\bar{z_2}}\), z₂ ≠ 0
|\(\bar{z}\)| = |z|
Z\(\bar{z}\) = |z|²
z ^-1 = \(\frac{\bar{z}}{|z|^2}\), z ≠ 0

Read More: sin cos tan values

Modulus of a Complex Number

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Assuming that z = x + iy is a complex number. Then, the positive square root of the total of the square of the real part and the square of the imaginary part is referred to as the modulus (absolute values) of z and it is represented by |z| i.e., |z| = x² + y²

It represents a distance of z from the origin in the set of complex number c, the order relation is not defined i.e., z₁ > z₂ or z₁ < z₂ doesn’t have any meaning but |z₁|<|z₂| or |z₁| > |z₂| has got its meaning, since |z₁| and |z₂| are real numbers.

De Moivre’s Theorem

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De Moivre's theorem generalizes the link to show that to raise a complex number to the nth power, increase the absolute value to the nth power and multiply the argument by n.

De Moivre’s Theorem

Let z = r(cos(Θ) + isin(Θ)

Then zⁿ = [r(cos(Θ)+isin(Θ)] ⁿ

= rⁿ(cos(nΘ) + isin(nΘ)

Where n is any positive integer.

Read More: Inverse Trigonometric Formulas

Proof of De Moivre’s Theorem

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De Moivre’s theorem states that (cosΘ+isinΘ)ⁿ = cos(nΘ) + isin(nΘ)

Assuming n = 1

(cosΘ+isinΘ)¹ = cos(nΘ) + isin(1Θ)

Assume n = k is true

So, (cosΘ+isinΘ) ^k = cos(nΘ) + isin(kΘ)

Letting n = k + 1

(cosΘ+isinΘ) (k+1) = cos(nΘ) + isin((k+1) Θ)

Assuming n = k, obtained equation:

= (cos(kΘ) + isin(kΘ)) x (cosΘ+isinΘ)

= cos(kΘ) cos(Θ)+icos(kΘ) sin(Θ)+isin(kΘ) cos(Θ)–sin(kΘ) sin(Θ)

Now previously known that,

sin(a+b) = sin(a)cos(b)+sin(b)cos(a)

and

cos(a+b) = cos(a)cos(b)–sin(a)sin(b)

= cos(kΘ+Θ) + isin(kΘ+Θ)

= cos((k+1) Θ) + isin((k+1) Θ)

Also Read:

Related Articles
Binomial expression	Roots	Difference Between Fraction and Rational Numbers

Solved Examples

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Example 1. Express the given complex number in the form a + ib: (1 – i)⁴

Solution: (1 − 1)² = [(1 − i)²] ²

= [1² + i² - 2i] ²

= [1 − 1 − 2i] ²

=(-2i)²

= (-2i) × (−2i)

= 4i² = 2=-4, [·.· i² =-1]

= a + ib, where a = – 4, and b = 0

Example 2. Express the given complex number 3(8 + i7) + i (8 + i7) in the form a + ib.

Solution: 3(8+i7) + i (8+ i7) = 24 + i21 + i8 + i × i8

As i x i = -1

3 (8+i7) + i (8+i7) = 24 + i21 + i8-8

3(8+i7) + i (8+ i7) = 16 + i29 = 16 + 29i

Things to Remember

A complex number is a number with both real and imaginary parts that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary part, satisfying the equation i² = −1.
Addition of Complex Numbers: (a+bi) +(c+di) = (a+c) + (b+d) i.
Subtraction of Complex Numbers: (a+bi) −(c+di) = (a−c) + (b−d) i.
Multiplication of Complex Numbers: (a+bi) ×(c+di) = (ac−bd) + (ad+bc) i
Multiplication Conjugates: (a+bi) × (a+bi) = a²+b²
Division of Complex Numbers: (a + ib) / (c + id) = (ac+bd)/ (c² + d²) + i(bc – ad) / (c² + d²).
Modulus of a Complex Number: |z| = x²+ y²
De Moivre's theorem generalizes the link to show that to raise a complex number to the nth power, increase the absolute value to the nth power and multiply the argument by n.

Sample Questions

Ques. Simplify 7i + 11i(3-i) (3 Marks)

Ans. 7i + 11i(3-i)

= 7i + 11i (3) + 11i (-i)

= 7i +33i – 11 i2

= 40 i – 11 (-1)

= 40i + 11

Ques. Write the given complex number (2 – i) – (–2 + i6) in the form a + ib (2 Marks)

Ans. Given Complex number: (2 – i) – (–2 + i6)

Multiply the sign (-) by the term inside the 2nd bracket (–1 + i6)

= 2 – i +2 – i6

= 4 – 7i, which is of the form a + ib.

Ques. Describe the mentioned complex number (-4) in the polar form. (3 Marks)

Ans. Given, the complex number is -3.

Assume r cos θ = -4 … (1) and r sin θ = 0 … (2)

When we square and add (1) and (2), we will have:

r²cos²θ + r²sin²θ = (-4)²

Take r² outside from L.H.S, we get

r²(cos²θ + sin²θ) = 16

We understand that, (cos²θ + sin²θ = 1), then the above-mentioned equation will become,

r² = 16

r = 4 (Conventionally, r > 0)

Now, putting the value of r in (1) and (2)

4 cos θ = -4, while 4 sin θ = 0

cos θ = -1 while sin θ = 0

Therefore, θ = π

Hence, the polar representation is,

-4 = r cos θ + i r sin θ

4 cos π + 4 sin π = 4(cos π + i sin π)

Thus, the required polar form is 4 cos π + 4i sin π = 4(cos π + i sin π).

Ques. Solve the given quadratic equation 3x²+ 2x + 1 = 0. (3 Marks)

Ans. Given the quadratic equation: 2x² + x + 1 = 0

Now, compare the given quadratic equation with the general form ax² + bx + c = 0

On comparing, we get

a = 3,

b = 2 and

c = 1

Therefore, the discriminant of the equation is:

D = b²– 4ac

Now, placing the values in the formula mentioned above

D = (2)² – 4(3)(1)

D = 4 - 12

D = -8

Thus, the asked solution for the provided quadratic equation will be:

x = [-b ± √D]/2a

x = [-2 ± √-8]/2(3)

We know that, √-1 = i

x = [-2 ± √8i] / 6

Hence, the solution for the given quadratic equation is (-2 ± √8i) / 6.

Ques. Prove that Re(z₁z₂) = Rez₁ Rez₂– Imz₁Imz₂ for any two complex numbers z₁ and z₂.. (3 Marks)

Ans. Provided: the two complex numbers i.e., z₁ and z₂

To prove: Re (z₁z₂) = Rez₁ Rez₂ – Imz₁ Imz₂

Assume z₁ = x₁ + iy₁ and

z₂ = x₂ + iy₂

Now, z₁ z₂ = (x₁ + iy₁) (x₂ + iy₂)

Now, separate the real part and the imaginary part from the formula mentioned above:

⇒ x₁ (x₂ + iy₂) + iy₁ (x₂ + iy₂)

Now, multiply the terms:

= (x₁×x₂) + (i×x₁×y₂) + (i×x₂×y₁) + (i²×y₁×y₂)

We are aware that, i² = -1, then we will have

=(x₁×x₂) + (i×x₁×y₂) + (i×x₂×y₁) – (i²×y₁×y₂)

Now, split the imaginary and the real part likewise:

= [(x₁×x₂) – (y₁×y₂)] + i [(x₁×y₂) + (x₂×y₁)]

Consider only the real part from the equation mentioned above:

⇒ Re (z₁×z₂) = (x₁×x₂) – (y₁×y₂)

It means that,

⇒ Re (z₁z₂) = Rez₁ Rez₂ – Imz₁ Imz₂

Hence, the given statement is proved.

Ques. Find the modulus of [(2+i)/(2-i)] – [(2-i)/(2+i)]. (3 Marks)

Ans. Given: [(2+i)/(2-i)] – [(2-i)/(2+i)]

Simplify the given expression, and we get:

[(2+i)/(2-i)] – [(2-i)/(2+i)] = [(2+i)2– (2-i)2]/ [(2+i) (2-i)]

= (2+i2+4i-2-i2+4i)) / (22+22)

Now, cancel out the terms,

= 8i/8

= 1i

Now, take the modulus,

| [(2+i)/(2-i)] – [(2-i)/(2+i)] | =|1i| = √12 = 1

Therefore, the modulus of [(2+i)/(2-i)] – [(2-i)/(2+i)] is 1.

Ques. Solve for x and y, 4x + (3x-y) i = 12 – 4i (2 Marks)

Ans. 4x = 12

x = 3

3x – y = - 4

3(3) – y = - 4

- y = - 4 – 9

y = 13

Ques. Write the real and imaginary parts 2 – 3i² (2 Marks)

Ans. Let z = 2 – 3i²

= 2 – 3 (-1)

= 2 + 3

= 5

= 5 + 0.i

Re (z) = 5, and

Im (z) = 0

Ques. For what real value of x and y are numbers equal (1+i) y² + (6+i) and (2+i) x? (2 Marks)

Ans. (1+i) × y² + (6 + i) = (2 + i) × x

y² + (i×y²) + 6 + i = 2x + xi

(y² + 6) + (y² + 1) i = 2x + xi

y² + 6 = 2x

y² + 1 = x

y² = x – 1

x – 1 + 6 = 2x

5 = x

y = ± 2

Ques. Find the real values of x and y if (x - iy) (3 + 5i) is the conjugate of – 6 + 24i (3 Marks)

Ans. (x – (i×y)) (3 + (5×i)) = - 6 + 24i

3x + 5xi – 3yi – 5yi² = - 6 + 24i

(3x + 5y) + (5x – 3y) i = - 6 + 24i

3x + 5y = -6

5x – 3y = 24

x = 3

y = -3

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Complex Number Formulas: Properties & De Moivre’s Theorem

What is a Complex Number?

Complex Numbers Detailed Video Explanation:

Equality of Complex Number Formula

Powers of Complex Numbers

Terms Used in Complex Numbers

Conjugate of Complex Number

Properties of Conjugate

Modulus of a Complex Number

De Moivre’s Theorem

Proof of De Moivre’s Theorem

Solved Examples

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

3.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

6.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

Similar Mathematics Concepts

Comments

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Complex Number Formulas: Properties & De Moivre’s Theorem

What is a Complex Number?

Complex Numbers Detailed Video Explanation:

Equality of Complex Number Formula

Powers of Complex Numbers

Terms Used in Complex Numbers

Conjugate of Complex Number

Properties of Conjugate

Modulus of a Complex Number

De Moivre’s Theorem

Proof of De Moivre’s Theorem

Solved Examples

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1. Find: If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

3.Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

4. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

6.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

3.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

6.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).