How do you convert $9 (\sqrt{3}) + 9 i$ $9(sqrt 3) + 9i$ to polar form?

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How do you convert $9 (\sqrt{3}) + 9 i$ $9(sqrt 3) + 9i$ to polar form?

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Answer 1 · 2016-04-25T04:42:09

$18 e^{\frac{π}{3} i}$ $18 e^(pi/3i)$

Explanation:

Here $z = (x, y) = (9 \sqrt{3}, 9) = (r cos θ, r sin θ)$ $z = (x, y) = (9 sqrt 3, 9)=(rcos theta, r sin theta )$ .

$r = \sqrt{x^{2} + y^{2}} = \sqrt{{(9 \sqrt{3})}^{2} + 9^{2}} = \sqrt{(9^{2}) (2^{2})} = 18$ $r=sqrt(x^2+y^2)=sqrt ((9sqrt 3)^2+9^2)=sqrt((9^2)(2^2))=18$ .
$cos θ = \frac{x}{r} = (9 \frac{\sqrt{3}}{18}) = \frac{\sqrt{3}}{2} and sin θ = \frac{y}{r} = \frac{9}{18} = \frac{1}{2}, and s o, θ = \frac{π}{3}$ $cos theta=x/r=(9sqrt 3/18)=sqrt 3/2 and sin theta = y/r=9/18=1/2, and so, theta=pi/3$ .

So, $z = (18, \frac{π}{3}) = 18 e^{\frac{π}{3} i}$ $z=(18, pi/3)=18 e^(pi/3i)$ .

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How do you convert $9 (\sqrt{3}) + 9 i$ $9(sqrt 3) + 9i$ to polar form?

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