Using De Moivre's Theorem, What is the indicated power of ${(- \sqrt{2} - \sqrt{2} i)}^{5}$ $(-sqrt2 -sqrt2 i)^5$ ?

Question

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

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Answer 1 · 2017-09-12T11:28:29

${(- \sqrt{2} - i \sqrt{2})}^{5}$ $(-sqrt2-isqrt2)^5$

$= {(2 (- \frac{1}{\sqrt{2}} - i \times \frac{1}{\sqrt{2}}))}^{5}$ $=(2(-1/sqrt2-ixx1/sqrt2))^5$

$= (2 {(- cos (\frac{π}{4}) - i \times sin (\frac{π}{4}))}^{5}$ $=(2(-cos(pi/4)-ixxsin(pi/4))^5$

$= (2 {(cos (π + \frac{π}{4}) + i \times sin (π + \frac{π}{4}))}^{5}$ $=(2(cos(pi+pi/4)+ixxsin(pi+pi/4))^5$

$= (2 {(cos (\frac{5 π}{4}) + i \times sin (\frac{5 π}{4}))}^{5}$ $=(2(cos((5pi)/4)+ixxsin((5pi)/4))^5$

$= (2^{5} (cos (\frac{25 π}{4}) + i \times sin (\frac{25 π}{4}))$ $=(2^5(cos((25pi)/4)+ixxsin((25pi)/4))$

$= (32 (cos (6 π + \frac{π}{4}) + i \times sin (6 π + \frac{π}{4}))$ $=(32(cos(6pi+pi/4)+ixxsin(6pi+pi/4))$

$= (32 (cos (\frac{π}{4}) + i \times sin (\frac{π}{4}))$ $=(32(cos(pi/4)+ixxsin(pi/4))$

$= 32 (\frac{1}{\sqrt{2}} + i \times \frac{1}{\sqrt{2}})$ $=32(1/sqrt2+ixx1/sqrt2)$

$= 16 \sqrt{2} + 16 \sqrt{2} i$ $=16sqrt2+16sqrt2i$