What is arctan⁡(z1)±arctan⁡(z2) with z1,z2∈C On wikipedia, there is the following identity: arctan⁡(u)±arctan⁡(v)=arctan⁡(u±v1∓uv) However when...

The problem is that arctan is a multivalued function. If you want a specific function, you need to specify which branch you are using. For example, the principal branch has real part in ${\displaystyle (-\frac{\pi }{2},\frac{\pi }{2}]}$. Other branches will differ from that one by an integer multiple of ${\displaystyle \pi }$. So the correct results are
${\displaystyle \arctan \left(u\right)\pm \arctan \left(v\right)=\arctan \left(\frac{u\pm v}{1\mp uv}\right)+n\pi }$
where n is an integer. If you are using the principal branch, it is the integer needed to put the real part of the arctan on the right in the interval ${\displaystyle (-\frac{\pi }{2},\frac{\pi }{2}]}$