Using De Moivre's Theorem, What is the indicated power of $(-sqrt2 -sqrt2 i)^5$ ?

Question

Ok, so I've gotten to $z=32[cos((25pi)/4)+isin((25pi)/4)]$ , but my book says the answer is $16sqrt(2)+16sqrt(2)i$ , how do I get there?

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Answer 1 · 2017-09-12T11:28:29

$(-sqrt2-isqrt2)^5$

$=(2(-1/sqrt2-ixx1/sqrt2))^5$

$=(2(-cos(pi/4)-ixxsin(pi/4))^5$

$=(2(cos(pi+pi/4)+ixxsin(pi+pi/4))^5$

$=(2(cos((5pi)/4)+ixxsin((5pi)/4))^5$

$=(2^5(cos((25pi)/4)+ixxsin((25pi)/4))$

$=(32(cos(6pi+pi/4)+ixxsin(6pi+pi/4))$

$=(32(cos(pi/4)+ixxsin(pi/4))$

$=32(1/sqrt2+ixx1/sqrt2)$

$=16sqrt2+16sqrt2i$