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Complex Number

Site: Newgate University Minna - Elearning Platform
Course: Basic Mathematics
Book: Complex Number
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Date: Monday, 6 April 2026, 2:42 PM

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Solutions and Explanation

1. Find the range of xx if 2x−6=62x - 6 = 6

Solve for xx:

2x−6=62x - 6 = 6

Add 6 to both sides:

2x=122x = 12

Divide by 2:

x=6x = 6

So, the range of xx is x=6x = 6 (since this is a specific value, the range is just {6}).

2. Absolute value of -11

The absolute value of a number is its non-negative distance from zero. So:

∣−11∣=11|-11| = 11

Correct answer: (c) 11

Introduction to Complex Numbers

  • The equation x2+1=0x^2 + 1 = 0 has no real solution since x2=−1x^2 = -1 is impossible in the real number system.

  • To solve this, we define the imaginary unit ii as:

    i=−1i = \sqrt{-1}

  • Thus, the solutions to the equation are:

    x=±ix = \pm i

Powers of ii

The imaginary unit follows a cycle:

  1. i=−1i = \sqrt{-1}

  2. i2=−1i^2 = -1

  3. i3=i2⋅i=(−1)⋅i=−ii^3 = i^2 \cdot i = (-1) \cdot i = -i

  4. i4=(i2)2=(−1)2=1i^4 = (i^2)^2 = (-1)^2 = 1

Since i4=1i^4 = 1, the powers of ii repeat in cycles of four.

Complex Number System (C\mathbb{C})

A complex number is of the form:

z=x+iyz = x + iy

where:

  • xx is the real part: Re⁡(z)=x\operatorname{Re}(z) = x

  • yy is the imaginary part: Im⁡(z)=y\operatorname{Im}(z) = y

The set of all complex numbers is denoted by C\mathbb{C}.

The Argand Diagram

The Argand Diagram is a way to visualize complex numbers on a coordinate plane:

  • The horizontal axis (x-axis) represents the real part (xx).

  • The vertical axis (y-axis) represents the imaginary part (yy).

For example, the complex number z=3+4iz = 3 + 4i is plotted at the point (3,4)(3, 4) in the plane.


Table of contents

1. Algebra of the Complex Number System

Algebra of the Complex Number System

The complex number system follows specific algebraic operations similar to real numbers, with additional properties due to the imaginary unit i=−1i = \sqrt{-1}. Let's break them down:

1. Addition and Subtraction of Complex Numbers

Let z1=x1+iy1z_1 = x_1 + i y_1 and z2=x2+iy2z_2 = x_2 + i y_2 be two complex numbers.

  • Addition is defined as:

    z1+z2=(x1+iy1)+(x2+iy2)=(x1+x2)+i(y1+y2)z_1 + z_2 = (x_1 + i y_1) + (x_2 + i y_2) = (x_1 + x_2) + i(y_1 + y_2)

  • Subtraction follows similarly:

    z1−z2=(x1+iy1)−(x2+iy2)=(x1−x2)+i(y1−y2)z_1 - z_2 = (x_1 + i y_1) - (x_2 + i y_2) = (x_1 - x_2) + i(y_1 - y_2)

This means that complex addition and subtraction are done component-wise for real and imaginary parts.

2. Multiplication of Complex Numbers

Multiplication follows the distributive property:

z1⋅z2=(x1+iy1)⋅(x2+iy2)z_1 \cdot z_2 = (x_1 + i y_1) \cdot (x_2 + i y_2)

Expanding using the distributive law:

x1x2+ix1y2+ix2y1+i2y1y2x_1 x_2 + i x_1 y_2 + i x_2 y_1 + i^2 y_1 y_2

Since i2=−1i^2 = -1, we simplify to:

(x1x2−y1y2)+i(x1y2+x2y1)(x_1 x_2 - y_1 y_2) + i (x_1 y_2 + x_2 y_1)

Thus, multiplication of two complex numbers results in another complex number.

3. Conjugate of a Complex Number

The conjugate of a complex number z=x+iyz = x + i y is denoted as:

z‾=x−iy\overline{z} = x - i y

Multiplying a complex number by its conjugate gives:

z⋅z‾=(x+iy)(x−iy)z \cdot \overline{z} = (x + i y)(x - i y)

Expanding using the difference of squares:

x2−ixy+ixy−i2y2x^2 - i x y + i x y - i^2 y^2

Since i2=−1i^2 = -1, this simplifies to:

x2+y2x^2 + y^2

which is the modulus squared of zz:

z⋅z‾=∣z∣2z \cdot \overline{z} = |z|^2

4. Rationalization (Finding the Reciprocal of a Complex Number)

The reciprocal of a complex number zz is given by:

z−1=1zz^{-1} = \frac{1}{z}

To rationalize the denominator, multiply by the conjugate:

z−1=1z×z‾z‾=z‾∣z∣2z^{-1} = \frac{1}{z} \times \frac{\overline{z}}{\overline{z}} = \frac{\overline{z}}{|z|^2}

This means:

1z=x−iyx2+y2\frac{1}{z} = \frac{x - i y}{x^2 + y^2}

This is useful in division.

5. Quotient (Division) of Complex Numbers

To divide two complex numbers:

z1z2=x1+iy1x2+iy2\frac{z_1}{z_2} = \frac{x_1 + i y_1}{x_2 + i y_2}

Multiply by the conjugate of the denominator:

z1z2=(x1+iy1)(x2−iy2)(x2+iy2)(x2−iy2)\frac{z_1}{z_2} = \frac{(x_1 + i y_1)(x_2 - i y_2)}{(x_2 + i y_2)(x_2 - i y_2)}

Since the denominator simplifies to ∣z2∣2=x22+y22|z_2|^2 = x_2^2 + y_2^2, we get:

z1z2=x1x2+y1y2∣z2∣2+iy1x2−x1y2∣z2∣2\frac{z_1}{z_2} = \frac{x_1 x_2 + y_1 y_2}{|z_2|^2} + i \frac{y_1 x_2 - x_1 y_2}{|z_2|^2}

This expresses the quotient in standard form.

6. Polar Form of a Complex Number

From Figure (3.1) (which you referenced), a complex number can also be expressed in polar form using:

x=rcos⁡θ,y=rsin⁡θx = r \cos\theta, \quad y = r \sin\theta

where:

  • r=∣z∣=x2+y2r = |z| = \sqrt{x^2 + y^2} is the modulus (magnitude) of zz.

  • θ=tan⁡−1(yx)\theta = \tan^{-1} \left(\frac{y}{x}\right) is the argument (angle) of zz.

Thus, we can write a complex number as:

z=r(cos⁡θ+isin⁡θ)z = r (\cos\theta + i\sin\theta)

This is called the trigonometric (polar) form of a complex number.


2. De Moivre’s Theorem

De Moivre’s Theorem

De Moivre’s theorem is a fundamental result in complex number theory that expresses the power of a complex number in polar form:

(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i\sin(n\theta)

where nn is any integer (positive, negative, or zero).

Proof by Mathematical Induction

Base Case ( n=1n = 1 )

For n=1n = 1:

(cos⁡θ+isin⁡θ)1=cos⁡θ+isin⁡θ(\cos \theta + i \sin \theta)^1 = \cos \theta + i \sin \theta

which is trivially true.

Inductive Step

Assume that the theorem holds for n=kn = k, i.e.,

(cos⁡θ+isin⁡θ)k=cos⁡(kθ)+isin⁡(kθ)(\cos \theta + i \sin \theta)^k = \cos(k\theta) + i \sin(k\theta)

Now, for n=k+1n = k+1:

(cos⁡θ+isin⁡θ)k+1=(cos⁡θ+isin⁡θ)⋅(cos⁡kθ+isin⁡kθ)(\cos \theta + i \sin \theta)^{k+1} = (\cos \theta + i \sin \theta) \cdot (\cos k\theta + i \sin k\theta)

Expanding using the distributive property:

cos⁡θcos⁡kθ+icos⁡θsin⁡kθ+isin⁡θcos⁡kθ+i2sin⁡θsin⁡kθ\cos \theta \cos k\theta + i \cos \theta \sin k\theta + i \sin \theta \cos k\theta + i^2 \sin \theta \sin k\theta

Since i2=−1i^2 = -1, we simplify:

cos⁡θcos⁡kθ−sin⁡θsin⁡kθ+i(cos⁡θsin⁡kθ+sin⁡θcos⁡kθ)\cos \theta \cos k\theta - \sin \theta \sin k\theta + i (\cos \theta \sin k\theta + \sin \theta \cos k\theta)

Using the trigonometric identities:

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B

we get:

cos⁡((k+1)θ)+isin⁡((k+1)θ)\cos((k+1)\theta) + i \sin((k+1)\theta)

Thus, by induction, De Moivre’s theorem holds for all positive integers nn.

Extension to Negative Powers

For n=−pn = -p, where pp is positive:

(cos⁡θ+isin⁡θ)−p=1(cos⁡θ+isin⁡θ)p(\cos \theta + i \sin \theta)^{-p} = \frac{1}{(\cos \theta + i \sin \theta)^p}

Using the identity:

cos⁡(−pθ)=cos⁡(pθ),sin⁡(−pθ)=−sin⁡(pθ)\cos(-p\theta) = \cos(p\theta), \quad \sin(-p\theta) = -\sin(p\theta)

we obtain:

(cos⁡θ+isin⁡θ)−p=cos⁡(−pθ)+isin⁡(−pθ)=cos⁡(pθ)−isin⁡(pθ)(\cos \theta + i \sin \theta)^{-p} = \cos(-p\theta) + i\sin(-p\theta) = \cos(p\theta) - i\sin(p\theta)

Thus, De Moivre’s theorem is valid for all integer values of nn.

Applications of De Moivre’s Theorem

  1. Finding Powers of Complex Numbers

    • Example: Compute (1+i)4(1 + i)^4.

    • Convert to polar form:

      1+i=2(cos⁡45∘+isin⁡45∘)1 + i = \sqrt{2} (\cos 45^\circ + i \sin 45^\circ)

    • Apply De Moivre’s theorem:

      (1+i)4=(2)4(cos⁡180∘+isin⁡180∘)=4(−1+0i)=−4(1+i)^4 = (\sqrt{2})^4 (\cos 180^\circ + i \sin 180^\circ) = 4(-1 + 0i) = -4

  2. Finding Roots of Complex Numbers

    • To find the nnth roots of a complex number z=r(cos⁡θ+isin⁡θ)z = r (\cos \theta + i \sin \theta), use:

      z1/n=r1/n[cos⁡(θ+2kπn)+isin⁡(θ+2kπn)],k=0,1,2,…,n−1z^{1/n} = r^{1/n} \left[ \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right], \quad k = 0,1,2,\dots, n-1