# How do you write the trigonometric form of -2-2i?

Nov 7, 2016

The trigonometric form is $z = 2 \sqrt{2} \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)$

#### Explanation:

Let $z = - 2 - 2 i$
The modulus of z is $| z | = \sqrt{{2}^{2} + {2}^{2}} = 2 \sqrt{2}$
Then we rewrite z as $z = 2 \sqrt{2} \left(- \frac{2}{2 \sqrt{2}} - \frac{2}{2 \sqrt{2}} i\right)$
simplify $z = 2 \sqrt{2} \left(- \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\right)$
Comparing this to $z = r \left(\cos \theta + i \sin \theta\right)$ which is the trgonometric form.
So $\cos \theta = - \frac{\sqrt{2}}{2}$ and $\sin \theta = - \frac{\sqrt{2}}{2}$
$\therefore \theta$ is in the 3rd quadrant
$\theta = \frac{5 \pi}{4}$
So the trigonometric form is #z=2sqrt2(cos((5pi)/4)+isin((5pi)/4))

and the exponential form is $z = 2 \sqrt{2} {e}^{\frac{5 i \pi}{4}}$