Complex Numbers I

  • The Idea behind ‘Imaginary Numbers’

The set of natural numbers is

{\mathbb N = \{ 1,2,3,..... \}}

This set of numbers does not have a solution to equations of the form

{a = x -y, x \le y}

Thus, we defined the set of whole numbers as

{\mathbb W = \{ 0,1,2,3,..... \}}

and the set of integers as

{\mathbb Z = \{ ...., -3, -2, -1, 0, 1, 2 ,3 ....\}}

Now, we do not have solution to the equations of the kind {a = \frac {x}{y}}, when {x} is not completely divisible by {y}. Hence, we defined rational numbers. Their set is given by

{\mathbb Q = \frac {p}{q}, s.t. \ p,q \in \mathbb Z, q \ne 0}

Still, we do not have solutions to equations of the form {x^n=a}, when {a} is not a perfect power. Thus, we defined irrational numbers. Combining the rational and irrational numbers, we get the set of real numbers, {\mathbb R}.

However, when {x^n =a, a<0} and {n} is an even power, we do not have solutions to such equations in {\mathbb R}. Therefore, we define a new {unit}, which is equal to {\sqrt {-1}}. It is denoted by {i}. Given this, we can solve any equation.

This number is an imaginary number. Thus, {i^2= -1}.

  • Complex Numbers

A number of the form {x+iy}, where {x, y \in \mathbb R} and {i = \sqrt {-1}} is a complex number. The set of complex numbers is denoted by {\mathbb C}.

Defining the complex numbers completes the system of numbers; i.e. given an equation in a single variable, we are sure that its solution will always be there in {\mathbb C}.

{x} is the real part and {y} is the imaginary part.

  • Algebra of Complex Numbers

Consider 2 complex numbers {z_1 = x_1 + i y_1} and {z_2 = x_2 + i y_2}.

I) Equality : They are equal only when {x_1 = x_2} and {y_1 = y_2}

II) Addition and Subtraction : {z_1 \pm z_2 = (x_1 \pm x_2) + i (y_1 \pm y_2)}

III) Multiplication : {z_1 \times z_2 = (x_1 x_2 - y_1y_2)+i(x_1y_2 - x_1y_2)}

IV) Division : {\frac {z_1}{z_2}} is done by multiplying numerator and denominator by complex conjugate of {z_2}.

Complex conjugate of {z_2} is {x_2 - i y_2}.

V) Addition and multiplication are commutative and associative. Also, multiplication distributes over addition.

  • Representation

On {XY} plane, complex number {x+iy} is represented by a point whose coordinates are {(x,y)}. The diagram representing the complex numbers is known as the Argand diagram.



In polar coordinates, {r = \sqrt {x^2 + y^2}} and {\theta = tan^{-1} \big( \frac y x \big)}.

{r} is known as the absolute value (or modulus) and {\theta} is known as the amplitude or argument of the complex number.

Thus, {z = x+iy = r cos (\theta) + i r sin (\theta) = r[cos (\theta)+ i sin (\theta)] = r e^{i \theta}}

The exponential representation of {z} is due to Euler.

Since {tan (\theta)} is periodic, we will consider {- \pi < \theta \le \pi} as the principal value of argument.

Thus, {mod(z) = \sqrt {x^2+y^2}} and {arg(z) = \tan^{-1} \big( \frac y x \big)}, {Re(z) = x} and {Im(z)=y}.

Let’s solve the following problem :

Show that for any complex number {z},

{\left |\dfrac {z}{|z|}   \right | \le |arg  \ z|}

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  • De Moivre’s Theorem

For any real number {n},

{\big \{ cos (\theta)+ i sin (\theta) \big \}^n = cos (n \theta) + i sin (n \theta)}

  • Applications of de Movire’s Theorem

In general, if {a^n = b}, then {a = b^{1/n}}. We say that {a} is the {n}th root of {b}. Further, {n}th root of any number has {n} values (which may not be always real).

For example, {9^{1/2}} has 2 roots , {3} and {-3}.

The theorem can be used to get {n}th roots of any number. This can be done by considering the periodicity of trigonometric functions. We know that

{cos (\theta) = cos (2n \pi + \theta) \ and \ sin (\theta) = sin (2n \pi + \theta), n \in \mathbb Z, n > 0}


{\big \{ cos (\theta)+ i sin (\theta) \big \}^{1/q} = cos \Big( \frac {2n \pi + \theta}{q} \Big) + i sin \Big ( \frac {2n \pi + \theta}{q} \Big)}

By substituting successive values of {n} from {0,1,2} till {q-1}, we get {q} different values of {\big \{ cos (\theta)+ i sin (\theta) \big \}^{1/q}}

This trick can be used to obtain the roots of an equation.

  • Complex Cube Roots Of Unity

In general, if the highest power of the variable in an equation is {n}, there will be {n} roots. Consider the equation


Clearly, {x=1} is a root. The other 2 roots are given by solving the quadratic equation below:

{(x^3-1)= (x^3-1^3)= (x-1){(x^2+x+1)}=0}

It gives

{\omega_1 = \frac {-1 + i \sqrt 3}{2} \ and \ \frac {-1 - i \sqrt 3}{2}}

Note, {\omega_1^2 = \omega_2} and {\omega_2^2 = \omega_1} and {1+\omega + \omega^2 = 1}.

When plotted on Argand diagram, the cube roots of unity produce an equilateral triangle, with {(1,0) \equiv 1+0i} as a vertex.

In general, if we are finding {n}th roots of unity, we will get a regular polygon of {n} sides and {(1,0)} will always be a vertex. This is because {x=1} is always a root of {x^n-1=0}.

Solved Example : Use de Moivre’s theorem to solve the equation {x^7 + x^4 + i (x^3+1)=0}.

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