# How do you convert 3sqrt(3) + 3i to polar form?

May 26, 2016

The result is $\left(6 , 0.52\right)$.

#### Explanation:

To convert a number from complex form to polar form you have simply to apply the definitions of sin and cos.

First of all we use the real part of the number ($3 \sqrt{3}$) as the $x$ coordinates and the imaginary part (without the $i$, so only $3$) as the $y$ and we put this on two axis as in the figure.

The polar form is nothing but $r$ and $\theta$.
$r$ can be obtained easily applying the Pitagora's theorem:

$r = \sqrt{{\left(3 \sqrt{3}\right)}^{2} + {3}^{2}} = \sqrt{27 + 9} = \sqrt{36} = 6$.

$\theta$ can be calculated using the definition of cosine or sine.
We know that $x = r \cos \left(\theta\right)$ and $y = r \sin \left(\theta\right)$. We could then say, for example, that $\sin \left(\theta\right) = \frac{y}{r}$ and then $\theta = \arcsin \left(\frac{y}{r}\right)$.
In the same way we can say $\theta = \arccos \left(\frac{x}{r}\right)$. But my favorite consists in dividing the two coordinates having

$\frac{y}{x} = \sin \frac{\theta}{\cos} \left(\theta\right)$ and because $\sin \frac{\theta}{\cos} \left(\theta\right) = \tan \left(\theta\right)$ we have $\frac{y}{x} = \tan \left(\theta\right)$ and finally

$\theta = \arctan \left(\frac{y}{x}\right)$. In our case $\theta = \arctan \left(\frac{3}{3 \sqrt{3}}\right) = \arctan \left(\frac{1}{\sqrt{3}}\right) \setminus \approx 0.52$.
Then your polar coordinates are $\left(6 , 0.52\right)$.
To see if the result is correct, you can transform back your number as:

$r \cos \left(\theta\right) + i r \sin \left(\theta\right) = 6 \cos \left(0.52\right) + i \cdot 6 \sin \left(0.52\right) = 5.19 + 3 i$ and 5.19 is the approximation of $3 \sqrt{3}$.