How do you use DeMoivre's Theorem to simplify #(3-2i)^5#?

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Answer 1 · 2017-02-04T06:45:18

#(3-2i)^5=69sqrt13(cos5alpha+isin5alpha)#,

where #alpha=tan^(-1)(-2/3)#

DeMoivre's Theorem states that if a complex number #z# in polar form is given by #z=r(costheta+isintheta)#

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

Here #|3-2i|=sqrt(3^2+2^2)=sqrt13#, hence we can write
#3-2i# as #sqrt13(cosalpha+isinalpha)#, where #alpha=tan^(-1)(-2/3)#

Hence #(3-2i)^5#

= #(sqrt13)^5(cos5alpha+isin5alpha)#

= #169sqrt13(cos5alpha+isin5alpha)#