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))#