# How do you convert 3+2i to polar form?

Sep 8, 2016

$\left(\sqrt{13} , 0.588\right)$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{cartesian to polar form}}$

That is $\left(x , y\right) \to \left(r , \theta\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here x = 3 and y = 2

$\Rightarrow r = \sqrt{{3}^{2} + {2}^{2}} = \sqrt{13}$

Now, 3 + 2i, is in the 1st quadrant so we must ensure that $\theta$ is in the 1st quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{2}{3}\right) = 0.588 \text{ radians"larr" in 1st quadrant}$

Thus $\left(3 , 2\right) \to \left(\sqrt{13} , 0.588\right) \to \left(\sqrt{13} , {33.69}^{\circ}\right)$