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

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

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Answer 1 · 2016-05-04T13:43:00

$\sqrt{3} + i$ $sqrt3+i$ in polar form is written as $2 cos (\frac{π}{6}) + 2 i sin (\frac{π}{6})$ $2cos(pi/6)+2isin(pi/6)$ or $2 e^{i (\frac{π}{6})}$ $2e^(i(pi/6)$

Explanation:

Complex number $a + i b$ $a+ib$ is written as $r cos θ + i r sin θ$ $rcostheta+irsintheta$ or $r e^{i θ}$ $re^(itheta)$ in polar form.

where, $r = \sqrt{a^{2} + b^{2}} 3 and$ $r=sqrt(a^2+b^2)3 and$ tantheta=b/a $or$ $or$ theta=arctan(b/a)#

Hence for $\sqrt{3} + i$ $sqrt3+i$

$r = \sqrt{{(\sqrt{3})}^{2} + 1^{2}} = \sqrt{4} = 2$ $r=sqrt((sqrt3)^2+1^2)=sqrt4=2$ and $θ = arctan (\frac{1}{\sqrt{3}}) = \frac{π}{6}$ $theta=arctan(1/sqrt3)=pi/6$

Hence $\sqrt{3} + i$ $sqrt3+i$ in polar form is written as $2 cos (\frac{π}{6}) + 2 i sin (\frac{π}{6})$ $2cos(pi/6)+2isin(pi/6)$

or $2 e^{i (\frac{π}{6})}$ $2e^(i(pi/6)$

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

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