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

Feb 14, 2015

You may use the de Moivre's Theorem:

$z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right] \implies {z}^{n} = {r}^{n} \left[\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right]$

z^4=2^4[cos(4*135°)+isin(4*135°)]=
=16[cos(540°)+isin(540°)]
(which is $= - 16$ in rectangular form)