How do you convert $1 + i \sqrt{3}$ $1 + i sqrt(3)$ to polar form?

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How do you convert $1 + i \sqrt{3}$ $1 + i sqrt(3)$ to polar form?

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Answer 1 · 2016-06-07T15:39:10

$(2, \frac{π}{3})$ $(2,pi/3)$

Explanation:

Using the formulae that link Cartesian to Polar coordinates.

$∙ r = \sqrt{x^{2} + y^{2}}$ $•r=sqrt(x^2+y^2)$

$∙ θ = {tan}^{- 1} (\frac{y}{x})$ $•theta=tan^-1(y/x)$

here x = 1 and y $= \sqrt{3}$ $=sqrt3$

$\Rightarrow r = \sqrt{1^{2} + {(\sqrt{3})}^{2}} = \sqrt{4} = 2$ $rArrr=sqrt(1^2+(sqrt3)^2)=sqrt4=2$

and $θ = {tan}^{- 1} (\sqrt{3}) = \frac{π}{3}$ $theta=tan^-1(sqrt3)=pi/3$

Thus $(1, \sqrt{3}) = (2, \frac{π}{3}) in polar form$ $(1,sqrt3)=(2,pi/3)" in polar form"$

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How do you convert $1 + i \sqrt{3}$ $1 + i sqrt(3)$ to polar form?

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