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Trigonometric Form of Complex Numbers

The complex numbers were discovered during 16th century by an Italian mathematician Gerolamo Cardano who gave them the name "fictitious" initially. A number that is expressible on a complex plane and is represented in terms of z = a + bi, where a is called real unit and b is termed as imaginary unit; is known as a complex number.

The letter i
 is known as iota and is given by $i^{2}$ = $-1$. The complex plane on which a complex number is denoted, is a two dimensional plane having horizontal axis for real part and vertical axis for imaginary part. Thus, a point (x, y) in Cartesian plane is identified by a point (x + yi) in complex plane. The complex number with zero real part, (i.e. a = 0 then z = ib), is a purely imaginary complex number. On the other hand, if imaginary part is zero (i.e. b = 0, then z = a), then it would be no more a complex number, it is rather a real number.

The complex number have wide applications. They are applied in chemistry, physics, economics, biol
ogy, statistics, engineering etc. In this article, we are going to learn about trigonometric form of complex numbers. These number can be represented in trigonometric form. How? In this article, we shall learn this. We shall also see some examples on trigonometrical representation of complex numbers.

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Express Complex Number in Trigonometric Form

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The complex number can be represented in the form of trigonometric ratios. This can be understood in the following way.
Graph :
We may illustrate complex number in trigonometric form by means of the graph shown below :
Trigonometric Form of a Complex Numbers
Let us suppose that a point (a, b) or a + bi is located on a complex plane. The horizontal axis or real axis denote the real part "a" of complex number which is the projection of given point on real axis. Similarly, the vertical axis or imaginary axis represents "b", the complex part which is the projection of point on imaginary axis.

Let us suppose the angle formed by line joining point with origin be $\alpha$ and r = |z| = a + ib. Then, finding the values of a and b using trigonometric rules.
a = r cos $\alpha$
b = r sin $\alpha$
In this way, our complex number is written in terms of trigonometry as below :
z = a + ib = r cos $\alpha$ + r sin $\alpha$
r (cos $\alpha$ + sin $\alpha$)

Properties

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The main properties of trigonometric form of complex numbers are discussed below :

(i) $(cos\ \alpha + i\ sin\ \alpha)^{n} = cos\ n \alpha + i\ sin\ n \alpha$, provided that n is an integer.

(ii) If n is given as a rational number, then $(cos\ n\ \alpha + i\ sin\ n \alpha)$ is one of the values of $(cos\ \alpha + i\ sin\ \alpha)^{n}$.

(iii) $(cos\ \alpha + i\ sin\ \alpha)^{-n} = (cos\ n\ \alpha - i\ sin\ n\ \alpha)$, where n is an integer.

(iv) $(cos\ \alpha + i\ sin\ \alpha) (cos\ \alpha - i\ sin\ \alpha) = 1$, therefore we may depict following -

$(cos\ \alpha + i\ sin\ \alpha)$ = $\frac{1}{cos\ \alpha - i\ sin\ \alpha}$ and
$(cos\ \alpha - i\ sin\ \alpha)$ = $\frac{1}{cos\ \alpha + i\ sin\ \alpha}$

(v) $(cos\ \alpha - i\ sin\ \alpha)^{n}$ = $[\frac{1}{cos\ \alpha + i\ sin\ \alpha}]^{n}$ = $\frac{1}{(cos \alpha + i\ sin \alpha)^{n}}$

Examples

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Below are the problems based on trigonometric form of complex numbers.

Example 1) Simplify (sinx + i cosx )$^{n}$.

Solution : (sinx + i cosx)$^{n}$

= $[cos (\frac{\pi}{2} - x) + i\ sin(\frac{\pi}{2} - x)]^{n}$

= $cos \ n(\frac{\pi}{2} - x) + i\ sin\ n(\frac{\pi}{2} - x)$

Example 2) Find the value of $(\frac{\sqrt{3}}{2} + \frac{i}{2})^{12}$.

Solution : $(\frac{\sqrt{3}}{2} + \frac{i}{2})^{12}$

$(cos\ 30^{\circ} + i\ sin\ 30^{\circ})^{12}$

$cos\ 360^{\circ} + i\ sin\ 360^{\circ}$

= 1 + i . 0 = 1

Example 3) Simplify $\frac{(cos\ \theta - i\ sin\ \theta)^{7}}{(sin\ 2 \theta - i\ cos\ 2 \theta)^{4}}$
.
Solution : $(cos\ \theta - i\ sin\ \theta)^{7} = (cos\ \theta + i\ sin\ \theta)^{-7}$
and
$(sin\ 2 \theta - i\ cos\ 2 \theta)^{4}$
= $[-i(cos\ 2 \theta + i\ sin\ 2\ \theta)]^{4}$
= $(-i)^{4} (cos\ 2 \theta + i\ sin\ 2\ \theta)^{4}$
= 1. $(cos\ \theta + i\ sin\ \theta)^{8}$

Now,
$\frac{(cos\ \theta - i\ sin\ \theta)^{7}}{(sin\ 2 \theta - i\ cos\ 2 \theta)^{4}}$

= $\frac{(cos\ \theta + i\ sin\ \theta)^{-7}}{(cos \theta + i\ sin\ \theta)^{8}}$

$(cos\ \theta + i\ sin\ \theta)^{-15}$
= $cos 15 \theta - i sin 15 \theta$

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