How do you convert ${(2 - 2 i)}^{5}$ $(2-2i)^5$ to polar form?

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Answer 1 · 2016-10-24T07:18:56

For $2 - i 2$ $2-i2$ , the polar form $r e^{i θ}$ $re^(itheta)$ is given by

$(2, - 2) = r (cos θ, sin θ)$ $(2, -2)=r(cos theta, sin theta)$ .

Here,

$r = \sqrt{2^{2} + {(- 2)}^{2}} = 2 \sqrt{2}$ $r = sqrt(2^2+(-2)^2)=2sqrt2$ ,

$cos θ = \frac{2}{r} = \frac{1}{\sqrt{2}} > 0$ $cos theta =2/r=1/sqrt 2>0$ and

$sin θ = - \frac{2}{r} = - \frac{1}{\sqrt{2}} < 0$ $sin theta =-2/r=-1/sqrt2<0$ .

The $Q_{4}$ $Q_4$ $θ$ $theta$ is $- \frac{π}{4}$ $-pi/4$

So, the given expression

${(2 - i 2)}^{5}$ $(2-i2)^5$

$= {(2 \sqrt{2} e^{i (- \frac{π}{4})})}^{5}$ $=(2sqrt2 e^(i(-pi/4)))^5$

$= 64 \sqrt{2} e^{5 (- i \frac{π}{4})}$ $=64sqrt2e^(5(-ipi/4))$

$- 64 \sqrt{2} e^{i (- 2 π + \frac{3}{4} π)}$ $-64sqrt2e^(i(-2pi+3/4pi))$

$= 64 \sqrt{2} e^{- i 2 π} e^{i (\frac{3}{4} π)}$ $=64sqrt2e^(-i2pi)e^(i(3/4pi))$

$= 64 \sqrt{2} (cos (\frac{3}{4} π) + i sin (\frac{3}{4} π))$ $=64sqrt2(cos(3/4pi)+isin(3/4pi))$ , using $e^{- i 2 π} = 1$ $e^(-i2pi)=1$

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How do you convert (2−2i)5(2-2i)^5 to polar form?