How do you simplify ${[2 (cos 135 d e g r e e s + i sin 135 d e g r e e s)]}^{4}$ $[2(cos 135degrees+i sin 135 degrees)]^4$ ?

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Answer 1 · 2015-02-14T20:15:42

You may use the de Moivre's Theorem:

$z = r [cos (θ) + i sin (θ)] \Rightarrow z^{n} = r^{n} [cos (n θ) + i sin (n θ)]$ $z=r[cos(theta)+isin(theta)] =>z^n=r^n[cos(ntheta)+isin(ntheta)]$

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)

How do you simplify [2(cos 135degrees+i sin 135 degrees)]^4[2(cos135degrees+isin135degrees)]4[2(cos 135degrees+i sin 135 degrees)]^4?