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

Question

7900 views around the world

You can reuse this answer
Creative Commons License

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 $theta=tan^(-1)(-2/3))$

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