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

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Answer 1 · 2016-09-19T14:07:48

$(2-2i)/(-1-i)=2(cos(pi/2)+isin(pi/2))$

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

$2-2i=r_1(cosalpha+isinalpha)$ and $-1-i=r_2(cosbeta+isinbeta)$

Then $(2-2i)/(-1-i)$ is given by

$(r_1(cosalpha+isinalpha)*(r_2(cosbeta-isinbeta)))/(r_2(cosbeta+isinbeta)(r_2(cosbeta-isinbeta)))$ which when simplified becomes

$(r_1*r_2(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/(r_2^2(cos^2beta+sin^2beta))$ or

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

$z_1/z_2$ is given by $(r_1/r_2, (alpha-beta))$ ,

Now as $|2-2i|=sqrt(2^2+(-2)^2)=sqrt8=2sqrt2$ and $2-2i=2sqrt2(1/sqrt2-i/sqrt2)=(cos((7pi)/4)+isin((7pi)/4)$

and $|-1-i|=sqrt((-1)^2+(-1)^2)=sqrt2$ and $-1-i=sqrt2(-1/sqrt2-i/sqrt2)=(cos((5pi)/4)+isin((5pi)/4))$

Hence, $(2-2i)/(-1-i)=((2sqrt2)/sqrt2)*(cos((7pi)/4-(5pi)/4)+isin((7pi)/4-(5pi)/4))$

= $2(cos(pi/2)+isin(pi/2))$

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