Mistake when calculating the polar form of $-3\mathrm{i}$

Question

Given the complex number $x+y\mathrm{i} = -3\mathrm{i}$, I want to calculate the polar form $r\mathrm{e}^{\varphi\mathrm{i}}$ using this definition for $\varphi$:

$\varphi = \begin{cases} \arctan(\frac y x) &\text{if } x > 0, \\ \arctan(\frac y x) + \pi &\text{if } x < 0 \text{ and } y \ge 0, \\ \arctan(\frac y x) - \pi &\text{if } x < 0 \text{ and } y < 0, \\ +\frac{\pi}{2} &\text{if } x = 0 \text{ and } y > 0, \\ -\frac{\pi}{2} &\text{if } x = 0 \text{ and } y < 0. \end{cases}$

so, if I understand correctly, since $x=0$ and $y = -3 < 0$, $\varphi$ should be $-\frac{\pi}{2}$ in this case, yielding

$3\mathrm{e}^{-\frac{\pi}{2}\mathrm{i}}$

but instead, the correct answer seems to be

$3\mathrm{e}^{\frac{3\pi}{2}\mathrm{i}}$.

I cannot see where I made the mistake and would be happy if you could help me with showing why exactly (and where) I am wrong. Thanks in advance.

Answer 1

It is only a matter of definition for the principal argument.

It can be

• $[0,2\pi)$

or

• $[-\pi,\pi)$

Thus both the polar form represent the same complex number $-3i$.