How do you convert $4 - 4 i$ $4-4i$ to polar form?

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How do you convert $4 - 4 i$ $4-4i$ to polar form?

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Answer 1 · 2016-04-18T03:34:49

$4 \sqrt{2} e^{i \frac{7 o u}{4}} = 4 \sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))$ $4sqrt2 e^(i(7ou)/4)=4sqrt2(cos((7pi)/4)+i sin((7pi)/4))$

Explanation:

$x = 4 and y = - 4$ $x=4 and y=-4$ .
$x + i y = r (cos θ + i sin θ)$ $x+i y=r (cos theta + i sin theta)$

$r = \sqrt{x^{2} + y^{2}} = 4 \sqrt{2}$ $r=sqrt(x^2+y^2)=4 sqrt 2$ .

$cos θ = \frac{x}{r} = \frac{1}{\sqrt{2}}, sin θ = \frac{y}{r} = - \frac{1}{\sqrt{2}}$ $cos theta=x/r=1/sqrt 2, sin theta=y/r=-1/sqrt2$

sine is negative and cosine is positive.
So, $θ$ $theta$ is in the fourth quadrant,
$θ = 2 π - \frac{π}{4} = \frac{7 π}{4}$ $theta=2pi-pi/4=(7pi)/4$ .

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How do you convert $4 - 4 i$ $4-4i$ to polar form?

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