# How do you express the complex number in trigonometric form: -6i?

Apr 16, 2016

$- 6 i = 6 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

#### Explanation:

Trigonometric form is where

$a + b i = \left\mid z \right\mid \left(\cos x + i \sin x\right)$.

$\left\mid z \right\mid$ is given by Pythagoras', $\sqrt{{a}^{2} + {b}^{2}}$, using the values from $a + b i$.

$a + b i = 0 - 6 i$
$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{0}^{2} + {\left(- 6\right)}^{2}} = \sqrt{36} = 6$

Now you can divide both sides by $6$, so

$\frac{- 6 i}{6} = \frac{6 \left(\cos x + i \sin x\right)}{6}$
$- i = \cos x + i \sin x$

This gives us, obviously, that

$\cos x = 0$
$\sin x = - 1$

and so, superimposing the graphs and finding points that satisfy the simultaneous equations, At the point $x = 270$, the $\cos x$ graph is $0$ and the $\sin x$ graph is at $- 1$, which satisfies the equations. Being repeating patterns, obviously there will be other results, such as $x = - 450 , - 90 , 630 , 990$ etc.

Putting all of this back together,

$- 6 i = 6 \left(\cos 270 + i \sin 270\right)$

or

$- 6 i = 6 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$