Question

Convert
`\sqrt3 + i` to polar form

Answer 1

I think the answer is 2(cos(π6)+isin(π6))

Answer 2

`a~~\pm~~bi\implies r[\cos(\theta ) \pm i\sin(\theta )] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ √(3)~~ + ~~i\implies \stackrel{x}{√(3)}~~ + ~~\stackrel{y}{1} i\qquad \qquad \begin{cases} r=√(x^2+y^2)\\\\ \theta =\tan^(-1)\left( (y)/(x) \right) \end{cases} \\\\[-0.35em] ~\dotfill`

`r=\sqrt{(√(3))^2+(1)^2}\implies r=√(3+1)\implies r=2 \\\\\\ \theta =\tan^(-1)\left( \cfrac{1}{√(3)} \right)\implies \theta =\cfrac{\pi }{6} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill 2\left[\cos\left( (\pi )/(6) \right)~~ + ~~i\sin\left( (\pi )/(6) \right) \right]~\hfill`

Check the picture below.