How do you use De Moivre's Theorem to simplify $z = {(2 [cos (5 \frac{π}{4}) + i sin (5 \frac{π}{4})])}^{5}$ $z=(2[cos(5pi/4)+isin(5pi/4)])^5$ to $16 \sqrt{2} + 16 \sqrt{2} i$ $16sqrt2 + 16sqrt2 i$ ?

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Answer 1 · 2016-12-10T13:03:32

$z = {(2 [cos (5 \frac{π}{4}) + i sin (5 \frac{π}{4})])}^{5}$ $z=(2[cos(5pi/4)+isin(5pi/4)])^5$

$\Rightarrow z = 2^{5} [cos (\frac{5 \times 5 π}{4}) + i sin (\frac{5 \times 5 π}{4})]$ $=>z=2^5[cos((5xx5pi)/4)+isin((5xx5pi)/4)]$

$\Rightarrow z = 2^{5} [cos (6 π + \frac{π}{4}) + i sin (6 π + \frac{π}{4})]$ $=>z=2^5[cos(6pi+pi/4)+isin(6pi+pi/4)]$

$\Rightarrow z = 2^{5} [cos (\frac{π}{4}) + i sin (\frac{π}{4})]$ $=>z=2^5[cos(pi/4)+isin(pi/4)]$

$\Rightarrow z = 32 [\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}]$ $=>z=32[1/sqrt2+i/sqrt2]$

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

How do you use De Moivre's Theorem to simplify z=(2[cos(5π4)+isin(5π4)])5z=(2[cos(5pi/4)+isin(5pi/4)])^5 to 16√2+16√2i16sqrt2 + 16sqrt2 i?