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#