complex numbers - Cartesian $-10i$ to Polar form

I am trying to convert the following problem to polar form:

$$z=-j10.$$

Using this equation, where $|z|=r=\sqrt{x^2+y^2}$ and $\arg z=\theta=\arctan(y/x).$

$$\eqalign{z&=|z|e^{j\arg z}\\ &=re^{j\theta}\\&=r\angle\theta.}$$

I determined that x = 0 and y = -10. However, if I plug x and y into arctan(y/x), the result would be indetermined since we're dividing by 0. The solution to that problem is 10<-90degrees.

Could someone give me some insight on how to convert the above cartesian to polar form?

Answer

From

http://hotmath.com/hotmath_help/topics/polar-form-of-a-complex-number.html

The polar of a complex number is given by:

$$z = r(\cos(\theta) + i\sin(\theta))$$
In your example:
$$z = -10i$$
$$r = \sqrt{0^2+(-10)^2} = 10$$
$$\theta = \arctan(\frac{-10}{0}) = \frac{3\pi}{2}$$
$\theta$ is $\frac{3\pi}{2}$ because the complex number is in the III quadrant. So the polar form of our complex number is $$z = 10(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}))$$