# Finding polar coordinates angle for complex numbers given cartesian form

I have the following formula for finding $$\theta$$ given cartesian form of complex numbers.

$$\theta = \begin{cases} \tan^{-1}(\frac{y}{x}) & x \leq 0 \\ \tan^{-1}(\frac{y}{x}) & x \geq 0 \\ \pm\frac{\pi}{2} & x = 0 \end{cases}$$

My confusion is when $$x=0$$. How do I tell if my angle is $$\frac{\pi}{2}$$ or $$-\frac{\pi}{2}$$

For example, let's say my complex number is $$z = -i$$

In that case $$x=0$$, so is the angle $$\frac{\pi}{2}$$ or $$-\frac{\pi}{2}$$?

• en.m.wikipedia.org/wiki/Atan2#Definition_and_computation Dec 16, 2018 at 3:08
• Are you sure the first two cases are identical? You might have a typo. Otherwise it would just be $x\ne 0$ Dec 16, 2018 at 3:15
• Dec 16, 2018 at 3:27

You need to try to understand the reason why that formula is given in the first place. The angle $$\theta$$ is the angle the complex number in the plane makes with the positive real axis (positive $$x$$-axis). Counting the angle in the positive direction is done counter clockwise. So if $$z = i$$, then it is clearly $$\displaystyle \frac{\pi}{2}$$ away from the positive $$x$$-axis purely from looking at the diagram.
In sum, if $$x = 0$$ then we have that $$z = iy$$ for some real number $$y$$. If $$y > 0$$ then $$\displaystyle \theta = \frac{\pi}{2}$$ and if $$y < 0$$ we have $$\displaystyle \theta = -\frac{\pi}{2}$$, try to convince yourself of this!