How do you use DeMoivre's Theorem to simplify #2(sqrt3+i)^7#?

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Answer 1 · 2016-10-17T10:29:02

#2(sqrt3+i)^7=-128sqrt3-128i#

According to DeMoivre's Theorem, if #z=r(costheta+isintheta)#

#z^n=r^n(cosntheta+isinntheta)#

Now #(sqrt3+i)#

= #2(sqrt3/2+ixx1/2#

= #2(cos(pi/6)+isin(pi/6))#

and using DeMoivre's Theorem, #(sqrt3+i)^7#

= #2^7(cos((7pi)/6)+isin((7pi)/6))#

= #128(cos(pi+pi/6)+isin(pi+pi/6))#

And #2(sqrt3+i)^7#

= #2xx128(cos(pi+pi/6)+isin(pi+pi/6))#

= #256(-cos(pi/6)-isin(pi/6))#

= #256(-sqrt3/2-i/2)#

= #-128sqrt3-128i#