How do you simplify #(- 1+ i ) ^ { 4}#?

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Answer 1 · 2017-10-05T17:52:51

The answer is #=-4#

Let the complex number be

#z=-1+i#

Convert this from the algebraic form to the tigonometric form

#z=|z|(-1/|z|+1/|z|i)#

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

So,

#z=sqrt2(-1/sqrt2+1/sqrt3i)#

Compare this to

#z=r(costheta+isintheta)#

#cos theta=-1/sqrt2# and #sintheta=1/sqrt2#

#theta=3/4pi#, #[mod2pi]#

#z=sqrt2(cos(3/4pi)+isin(sin(3/4pi))=sqrt2e^(3/4pii)#

Therefore,

#z^4=(sqrt2e^(3/4pii))^4=(sqrt2)^4e^(3pii)#

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

#=-4#