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How do you simplify #[2(cos 135degrees+i sin 135 degrees)]^4#?

1 Answer

Feb 14, 2015

You may use the de Moivre's Theorem:

#z=r[cos(theta)+isin(theta)] =>z^n=r^n[cos(ntheta)+isin(ntheta)]#

In your case you get:

#z^4=2^4[cos(4*135°)+isin(4*135°)]=#

#=16[cos(540°)+isin(540°)]#

(which is #=-16# in rectangular form)

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