<s>[INST] Complex integral $\int^{2\pi}_0 e^{i\tan^{-1}\frac{r\sin(t-t_0)}{1+r\cos(t-t_0)}-i\tan^{-1}\frac{1-r \cos( t-t_0)}{r\sin(t-t_0)}}dt$ Is there a way to evaluate the following integral by simplifying the exponent?
$$\int^{2\pi}_0 \exp \left(i\tan^{-1}\frac{r\sin(\phi-\phi_0)}{1+r\cos(\phi-\phi_0)}-i\tan^{-1}\frac{1-r \cos( \phi-\phi_0)}{r\sin(\phi-\phi_0)}\right) \, d\phi$$
where $r$ and $\phi_0$ are constants.
 [/INST] $$
\arctan u + \arctan v = \arctan \frac{u+v}{1-uv} \tag 1
$$
\begin{align}
& \arctan \frac A {1+B} - \arctan\frac{1-B} A = \arctan \frac{\frac A{1+B} - \frac{1-B} A}{1 + \frac{A(1-B)}{A(1+B)}} \\[10pt]
= {} & \arctan\frac{A^2 - (1-B^2)}{A(1+B) + A(1-B)} = \arctan \frac{A^2+B^2 - 1}{2A}
\end{align}
In your case $A^2+B^2 = r^2$ and $2A = 2r\sin(\varphi - \varphi_\xi).$
But you need to think through the question of when $(1)$ holds, since "arctan" is multiple-valued unless the domain of the tangent function is restricted.
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