How do I use DeMoivre's theorem to find #(2+2i)^6#?

1 Answer
May 10, 2018

Answer:

The answer is #=-512i#

Explanation:

Let #z=2+2i=2(1+i)#

Transform #z=a+ib# from algebraic form into polar form

#z=|z|(a/(|a|)+ib/(|z|))#

where,

#{(costheta=a/(|z|)),(sintheta=b/(|z|)):}#

Here,

#|z|=sqrt(1^2+1^2)=sqrt2#

#z=2sqrt2(1/sqrt2+i/sqrt2)#

#{(costheta=1/sqrt2),(sintheta=1/sqrt2)):}#

#=>#, #theta=1/4pi#, #[2pi]#

The polar form is

#z=2sqrt2(cos(pi/4)+isin(pi/4))#

Demoivre's theorem is

#(costheta+isintheta)^n=cos(ntheta)+isin(ntheta)#

Therefore,

#(2+2i)^6=(2sqrt2(cos(pi/4)+isin(pi/4)))^6#

#=(2sqrt2)^6(cos(pi/4*6)+isin(pi/4*6))#

#=(2sqrt2)^6(cos(3/2pi)+isin(3/2pi))#

#=512(0-i)#

#=-512i#