How do you convert $4 (\sqrt{3} + i)$ $4 (sqrt(3) +i)$ to polar form?

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How do you convert $4 (\sqrt{3} + i)$ $4 (sqrt(3) +i)$ to polar form?

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Answer 1 · 2016-04-18T03:48:29

$8 e^{i (\frac{π}{3})} = 8 (cos (\frac{π}{6}) + i sin (\frac{π}{6}))$ $8 e^(i(pi/3))=8 ( cos (pi/6) + i sin (pi/6))$

Explanation:

$x = 4 \sqrt{3} = y$ $x=4sqrt3=y$
$x + i y = r (cos θ + i sin θ)$ $x+ i y = r ( cos theta + i sin theta )$

$r = \sqrt{x^{2} + y^{2}} = 8$ $r=sqrt(x^2+y^2)=8$
$cos θ = \frac{x}{r} = \frac{\sqrt{3}}{2}, sin θ = \frac{y}{r} = \frac{1}{2}$ $cos theta = x/r = sqrt3/2, sin theta = y/r=1/2$ , Both are positive. $θ$ $theta$ is in the 1st quadrant.
$θ = \frac{π}{6}$ $theta=pi/6$
So, $4 (\sqrt{3} + i) = 8 (\frac{cos π}{6} + i \frac{sin π}{6})$ $4( sqrt3 + i )=8 (cos pi/6 + i sin pi/6 )$

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How do you convert $4 (\sqrt{3} + i)$ $4 (sqrt(3) +i)$ to polar form?

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