How do you find the power $[2 {(cos (\frac{π}{4}) + i sin (\frac{π}{4})]}^{5}$ $[2(cos(pi/4)+isin(pi/4)]^5$ and express the result in rectangular form?

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Answer 1 · 2017-02-22T07:08:17

${[2 (cos (\frac{5 π}{4}) + i sin (\frac{5 π}{4}))]}^{5} = - 16 \sqrt{2} - i 16 \sqrt{2}$ $[2(cos((5pi)/4)+isin((5pi)/4))]^5=-16sqrt2-i16sqrt2$

According to DeMoivre's theorem if $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

then $z^{n} = r^{n} (cos n θ + i sin n θ)$ $z^n=r^n(cosntheta+isinntheta)$

Hence, if $z = 2 (cos (\frac{π}{4}) + i sin (\frac{π}{4}))$ $z=2(cos(pi/4)+isin(pi/4))$

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

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

= $32 (- cos (\frac{π}{4}) - i sin (\frac{π}{4}))$ $32(-cos(pi/4)-isin(pi/4))$

= $32 (- \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}})$ $32(-1/sqrt2-i1/sqrt2)$

= $- \frac{32}{\sqrt{2}} - i \frac{32}{\sqrt{2}}$ $-32/sqrt2-i32/sqrt2$

= $- 16 \sqrt{2} - i 16 \sqrt{2}$ $-16sqrt2-i16sqrt2$

How do you find the power [2(cos(π4)+isin(π4)]5[2(cos(pi/4)+isin(pi/4)]^5 and express the result in rectangular form?