How do you use DeMoivre's Theorem to simplify #(3-2i)^8#?

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Answer 1 · 2017-01-02T06:57:46

#(3-2i)^8=13^4(cos8theta+isin8theta)#, where #theta=tan^(-1)(-2/3))#

According to DeMoivre's theorem, if #z=r(costheta+isintheta)#

then #z^n=r^n(cosntheta+isinntheta)#

Hence to simplify #(3-2i)^8#, we should first put #(3-2i)# in polar form.

As #rcostheta=3# and #rsintheta=-2#, #r=sqrt(3^2+2^2)=sqrt13#

and #theta=tan^(-1)(-2/3)#

Hence #(3-2i)^8=(sqrt13)^8(cos8theta+isin8theta)#

= #13^4(cos8theta+isin8theta)#, where #theta=tan^(-1)(-2/3))#