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 $(3-2i)^5$

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

= $169sqrt13(cos5alpha+isin5alpha)$

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