How does arctan(0) equal pi?

It is an oversimplification to say the argument of $x+iy$ is given by $\theta =\arctan \dfrac yx$? Actually you have to solve the system \begin{cases}\cos\theta=\dfrac x{\sqrt{x^2+y^2}}\\ \sin\theta=\dfrac y{\sqrt{x^2+y^2}}\end{cases} which implies $\tan\theta=\dfrac yx$, but is not equivalent to $\theta=\arctan\dfrac yx$.

The general solution is $\theta\equiv\arctan\dfrac yx\mod \pi$, and you have to choose the value of $\theta (mod $2\pi) in function of the signs of $\cos\theta$ and $\sin\theta$, i.e. of $x$ and $y$.

$\arctan(0)=0,$ not $\pi$, but $\tan(\pi)=\tan(0)=0$ As $x$ ranges from $-\frac \pi 2$ to $\frac \pi 2$, $\tan x$ ranges from $-\infty$ to $+\infty$. As we want $\arctan$ to be a function, we have to choose a range of output values. When you use the $\arctan$ to get the polar angle, there is an uncertainty of $\pi$ because $\tan (x+\pi)=\tan x$. You have to use the real and imaginary parts to decide the multiple of $\pi$ to add to the $\arctan$ output to get the proper polar angle you want. You might even need to add $2\pi$. For example, if you want the polar form of $1-i$, the $\arctan$ will give you $-\frac \pi 4$ but you probably want to report the polar angle as $\frac {7 \pi}4$

$\tan(x)=0$ has the solutions $x=n\pi$, so $\pi$ is also a solution. Which solution you choose, depends on the signs of $\sin(x)$ and $\cos(x)$. This is why computers have a special function atan2, which calculates the solutions of $\tan(x)=\frac{u}{v}$ and returns the value in the proper quadrant, by accounting for the signs of $u$ and $v$ (as well as for the values where $\tan(x)=\frac{u}{v}$ might be ambiguous/undefined).

because the angle of a complex number is not given by the ordinary artan, is necessary make an adjust of Pi if the real part is negative, add pi and ist done