How do you use demoivre's theorem to simplify $[2(cos((pi)/2)+isin((pi)/2))]^8$ ?

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Answer 1 · 2016-10-15T16:55:23

$[2(cos(pi/2)+isin(pi/2))]^8=256$

According to de Moivre's Theorem we can calculate any integer power of a complex number given in trigonometric form.

The theorem says that:

If a complex number $z$ is given in a form:

$z=|z|(cosvarphi+isinvarphi)$

Then the $n-th$ power of $z$ is:

$z^n=|z|^n(cosnvarphi+isinnvarphi)$

Using this formula we get:

$[2(cos(pi/2)+isin(pi/2))]^8=2^8[cos((8pi)/2)+isin((8pi)/2)]=$

$=256[cos4pi+isin4pi]=256$

How do you use demoivre's theorem to simplify [2(cos((pi)/2)+isin((pi)/2))]^8[2(cos((pi)/2)+isin((pi)/2))]^8?