How do you use DeMoivre's theorem to simplify #(-1+i)^4#?

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Answer 1 · 2017-08-21T05:46:34

The answer is #=-4#

We start by writing #z=-1+i#, in trigonometric form

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

#z=sqrt2(-1/sqrt2+1/sqrt2i)#

#costheta=-1/sqrt2#, #=>#, #theta=3/4pi#

#sintheta=1/sqrt2#, #=>#, #theta=3/4pi#

The trigonometric form is

#z=sqrt2(cos(3/4pi)+isin(3/4pi))#

Applying Demoivre's theorem

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

#z^4=(sqrt2(cos(3/4pi)+isin(3/4pi))^4#

#=(sqrt2)^4(cos(3/4pi*4)+isin(3/4pi*4))#

#=4(cos(3pi)+isin(3pi))#

#=4(-1+i*0)#

#=-4#