How do you use DeMoivre's Theorem to simplify $(2+2i)^6$ ?

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Answer 1 · 2016-10-07T12:58:12

$(2+2i)^6=-512i$

De Moivre's Theorem states that if a complex number

$z=r(costheta+isintheta)$ , then

$z*n=r^n(cosntheta+isinntheta)$

Now, in $2+2i$ as its absolute value is $sqrt(2^2+2^2)=sqrt8=2sqrt2$ , it can be written as

$2+2i=2sqrt2(1/sqrt2+i*1/sqrt2)$

= $2sqrt2(cos(pi/4)+isin(pi/4))$

Hence $(2+2i)^6=(2sqrt2)^6(cos(pi/4xx6)+isin(pi/4xx6))$

= $2^6xx2^3(cos((3pi)/2)+isin((3pi)/2))$

= $512(0-i)=-512i$

How do you use DeMoivre's Theorem to simplify (2+2i)^6(2+2i)^6?