What is $\arctan (z_1) \pm \arctan (z_2)$ with $z_1,z_2\in\mathbb{C}$

Question

On wikipedia, there is the following identity: $$\arctan (u) \pm \arctan (v)=\arctan \left(\frac{u \pm v}{1 \mp u v}\right)$$

However when I try some $u,v\in\mathbb{C}$ to check, the formula does not hold. Is there an equivalent formula for complex numbers?

Answer 1

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 $(-\pi/2, \pi/2]$. Other branches will differ from that one by an integer multiple of $\pi$. So the correct results are

$$\arctan(u) \pm \arctan(v) = \arctan\left( \frac{u\pm v}{1\mp uv} \right) + n \pi $$ 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 $(-\pi/2, \pi/2]$.

Answer 2

With a calculator,

$$\arctan(1+i)-\arctan(2-i)=-0.1608752771983+0.5756462732485i,$$

$$\arctan\frac{1+i-2+i}{1+(1+i)(2-i)}=\arctan\frac{-2+9i}{17}=-0.1608752771983+0.5756462732485i.$$

In other cases, you can have a discrepancy of $k\pi$.