# 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 ${z}$,

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

• ### 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

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

Then,

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

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

${x^3-1=0}$

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 ${x^7 + x^4 + i (x^3+1)=0}$.