How do you convert $2 + (\sqrt{3}) i$ $2+(sqrt 3)i$ to polar form?

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How do you convert $2 + (\sqrt{3}) i$ $2+(sqrt 3)i$ to polar form?

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Answer 1 · 2016-03-19T05:15:55

The polar coordinates are: $P (\sqrt{7}, .714)$ $P(sqrt(7),.714)$

Explanation:

calculate the magnitude, r:
$r = \sqrt{2^{2} + {(\sqrt{3})}^{2}} = \sqrt{4 + 3} = \sqrt{7}$ $r = sqrt(2^2+(sqrt(3))^2)=sqrt(4+3)=sqrt(7)$
Calculate the angle $θ$ $theta$ :
$θ = {tan}^{- 1} (\frac{\sqrt{3}}{2}) = {40.89}^{0} = .714 radians$ $theta=tan^-1 (sqrt(3)/2)=40.89^0=.714 " radians"$
The polar coordinates are: $P (\sqrt{7}, .714)$ $P(sqrt(7),.714)$

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How do you convert $2 + (\sqrt{3}) i$ $2+(sqrt 3)i$ to polar form?

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