# How do you convert (2-2i)/(-1-i) to polar form?

Sep 19, 2016

$\frac{2 - 2 i}{- 1 - i} = 2 \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

#### Explanation:

Let us write the two complex numbers in polar coordinates and let them be

$2 - 2 i = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and $- 1 - i = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

Then $\frac{2 - 2 i}{- 1 - i}$ is given by

$\frac{{r}_{1} \left(\cos \alpha + i \sin \alpha\right) \cdot \left({r}_{2} \left(\cos \beta - i \sin \beta\right)\right)}{{r}_{2} \left(\cos \beta + i \sin \beta\right) \left({r}_{2} \left(\cos \beta - i \sin \beta\right)\right)}$ which when simplified becomes

$\frac{{r}_{1} \cdot {r}_{2} \left(\cos \alpha \cos \beta + \sin \alpha \sin \beta\right) + i \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)}{{r}_{2}^{2} \left({\cos}^{2} \beta + {\sin}^{2} \beta\right)}$ or

(r_1/r_2)*(cos(alpha-beta)+isin(alpha-beta) or

${z}_{1} / {z}_{2}$ is given by $\left({r}_{1} / {r}_{2} , \left(\alpha - \beta\right)\right)$,

Now as $| 2 - 2 i | = \sqrt{{2}^{2} + {\left(- 2\right)}^{2}} = \sqrt{8} = 2 \sqrt{2}$ and 2-2i=2sqrt2(1/sqrt2-i/sqrt2)=(cos((7pi)/4)+isin((7pi)/4)

and $| - 1 - i | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{2}$ and $- 1 - i = \sqrt{2} \left(- \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)$

Hence, $\frac{2 - 2 i}{- 1 - i} = \left(\frac{2 \sqrt{2}}{\sqrt{2}}\right) \cdot \left(\cos \left(\frac{7 \pi}{4} - \frac{5 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4} - \frac{5 \pi}{4}\right)\right)$

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