What is the trigonometric form of  (-1+8i) ?

Feb 28, 2017

As explained below

Explanation:

If denoted by z = -1+8i, then

|z|= $\sqrt{{1}^{2} + {8}^{2}} = \sqrt{65}$

Now write z= $\sqrt{65} \left(- \frac{1}{\sqrt{65}} + i \frac{8}{\sqrt{65}}\right)$

Now consider an angle $\theta$, such that $\cos \theta = - \frac{1}{\sqrt{65}}$ and $\sin \theta = \frac{8}{\sqrt{65}}$. This implies $\tan \theta = \frac{8}{-} 1 = - 8$

Now z can be expressed as $\sqrt{65} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{-} 1 - 8 \mathmr{and} a r c \tan - 8$

This is the required trignometric form.