How do you convert $\frac{2 - 2 i}{- 1 - i}$ $(2-2i)/(-1-i)$ to polar form?

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How do you convert $\frac{2 - 2 i}{- 1 - i}$ $(2-2i)/(-1-i)$ to polar form?

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Answer 1 · 2016-09-19T14:07:48

$\frac{2 - 2 i}{- 1 - i} = 2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $(2-2i)/(-1-i)=2(cos(pi/2)+isin(pi/2))$

Explanation:

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

$2 - 2 i = r_{1} (cos α + i sin α)$ $2-2i=r_1(cosalpha+isinalpha)$ and $- 1 - i = r_{2} (cos β + i sin β)$ $-1-i=r_2(cosbeta+isinbeta)$

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

$\frac{r_{1} (cos α + i sin α) \cdot (r_{2} (cos β - i sin β))}{r_{2} (cos β + i sin β) (r_{2} (cos β - i sin β))}$ $(r_1(cosalpha+isinalpha)*(r_2(cosbeta-isinbeta)))/(r_2(cosbeta+isinbeta)(r_2(cosbeta-isinbeta)))$ which when simplified becomes

$\frac{r_{1} \cdot r_{2} (cos α cos β + sin α sin β) + i (sin α cos β - cos α sin β)}{r_{2}^{2} ({cos}^{2} β + {sin}^{2} β)}$ $(r_1*r_2(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/(r_2^2(cos^2beta+sin^2beta))$ or

$(\frac{r_{1}}{r_{2}}) \cdot (cos (α - β) + i sin (α - β)$ $(r_1/r_2)*(cos(alpha-beta)+isin(alpha-beta)$ or

$\frac{z_{1}}{z_{2}}$ $z_1/z_2$ is given by $(\frac{r_{1}}{r_{2}}, (α - β))$ $(r_1/r_2, (alpha-beta))$ ,

Now as $| 2 - 2 i | = \sqrt{2^{2} + {(- 2)}^{2}} = \sqrt{8} = 2 \sqrt{2}$ $|2-2i|=sqrt(2^2+(-2)^2)=sqrt8=2sqrt2$ and $2 - 2 i = 2 \sqrt{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4})$ $2-2i=2sqrt2(1/sqrt2-i/sqrt2)=(cos((7pi)/4)+isin((7pi)/4)$

and $| - 1 - i | = \sqrt{{(- 1)}^{2} + {(- 1)}^{2}} = \sqrt{2}$ $|-1-i|=sqrt((-1)^2+(-1)^2)=sqrt2$ and $- 1 - i = \sqrt{2} (- \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = (cos (\frac{5 π}{4}) + i sin (\frac{5 π}{4}))$ $-1-i=sqrt2(-1/sqrt2-i/sqrt2)=(cos((5pi)/4)+isin((5pi)/4))$

Hence, $\frac{2 - 2 i}{- 1 - i} = (\frac{2 \sqrt{2}}{\sqrt{2}}) \cdot (cos (\frac{7 π}{4} - \frac{5 π}{4}) + i sin (\frac{7 π}{4} - \frac{5 π}{4}))$ $(2-2i)/(-1-i)=((2sqrt2)/sqrt2)*(cos((7pi)/4-(5pi)/4)+isin((7pi)/4-(5pi)/4))$

= $2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $2(cos(pi/2)+isin(pi/2))$

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