The best way to determine the $arg(z)$ or inverse tangent of a value to find polar form of complex number

Question

First of all, this question is focused on how to find the $\arg(z)$ or inverse tangent of ratio of $\Im(z)$ and $\Re(z)$. And please to note that, before i'm asking this question, i've read all the possible duplicates. But all of those have many different answer and make me more confused.

Suppose i have $-1+0i$, the polar form is :

1. $r=|-1+0i|=1$
2. $\theta=\tan^{-1}(0)=\pi, \quad 0\,\text{is satisfied but why gives different answer?} $
3. $\therefore -1= \text{cis}(\pi)=e^{i\pi}$

Suppose i have $3+0i$, the polar form is :

1. $r=|3+0i|=3$
2. $\theta=\tan^{-1}(0)=0, \quad \pi \,\text{gives different answer?}$
3. $\therefore 3= 3\,\text{cis}(0)=3e^{0i}$

Suppose i have $0-4i$, the polar form is :

1. $r=|0-4i|=4$
2. $\theta=\tan^{-1}\left(\dfrac{-4}{0}\right)=-\dfrac{\pi}{2}, \quad \dfrac{\pi}{2} \,\text{gives different answer?}$
3. $\therefore -4i= 4\,\text{cis}\left(-\dfrac{\pi}{2}\right)=3e^{-i\frac{\pi}{2}}$

From those example, why in step $2$ for each example has different result?

Besides, i found a formula or something like that (on my possible duplicate), it said :

$$\theta=\arg(z)=\begin{cases} 2\tan^{-1}\left(\dfrac{\Im(z)}{|z|+\Re(x)}\right) &\text{if $x>0 \lor y\ne 0$}\\ \pi &\text{if $x<0 \land y=0$}\\ \text{Undefined} &\text{if $x=0 \land y=0$} \end{cases}$$

Is that formula always work in all cases? But it doesn't give me enough information when $x>0 \land y=0$ (is this includes the first rule? When $x>0 \lor y\ne 0$)

And many of them (on my possible duplicate) said the $\theta$ needs to add $\pi$ or $2k\pi,\,k\in \Bbb{N}$ (Which one is true?)

Thanks for advance.

Answer 1

the easy way to understand the disparities above is noting that the argument is multivalued and the tangent captures only half of that since it has period $\pi$ while the original $x,y$ are defined in terms of $\cos, \sin$ which have period $2\pi$ so when inverting the tangent you need to choose the one value that matches the real and imaginary parts - so for example in your first example the real part is negative and while correctly you identify $0, \pi$ as the two possibilities, you need to match the cosine too and only $\pi$ then works ; in example two, matching cosine gives you $0$, while in example 3 you need to match the sine and that gives you the right value out the two again