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 · 2016-06-23T13:01:10

I found: $2 ∠ 30^{\circ}$ $2/_30^@$

Explanation:

Have a look:

enter image source here

Answer 2 · 2016-06-23T13:05:01

$(2, \frac{π}{6})$ $(2,pi/6)$

Explanation:

To convert cartesian coordinates (x ,y) to Polar coordinates $(r, θ)$ $(r,theta)$
Use the formulae which link them.

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

and since $\sqrt{3} + i = (\sqrt{3}, 1)$ $sqrt3+i=(sqrt3,1)$ is in the 1st quadrant , then.

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

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

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

and $θ = {tan}^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6} or 30^{\circ}$ $theta=tan^-1(1/sqrt3)=pi/6" or "30^@$

$\Rightarrow (\sqrt{3}, 1) \to (2, \frac{π}{6})$ $rArr(sqrt3,1)to(2,pi/6)$

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

2 Answers

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How do you convert √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?