I would like to understand if the result of following integral
$$A(\alpha)=\int_0^{2\pi}\,d\theta\,\,\frac{a-i\cdot{b}\cos(\pi+\theta+\alpha)}{c-i\cdot{d}\cos(\pi+\theta+\alpha)}$$
is or not dependent on the value $\alpha\in[0 \,\,2\pi]$ (with $a,b,c,d\in\mathbf{R}^+$).
As far as I understood from one of my previous questions is that the integral must be solved using the residue theorem.
For the sake of tidiness, note first that $\frac{a-i\cdot{b}\cos(\pi+\theta+\alpha)}{c-i\cdot{d}\cos(\pi+\theta+\alpha)}=\frac{a+i\cdot{b}\cos(\theta+\alpha)}{c+i\cdot{d}\cos(\theta+\alpha)}.$
Substituting $\phi=\theta+\alpha$, $$A(\alpha) = \int_0^{2\pi}\,d\theta\,\,\frac{a+i\cdot{b}\cos(\theta+\alpha)}{c+i\cdot{d}\cos(\theta+\alpha)} = \int_{\alpha}^{2\pi+\alpha}\,d\phi\,\,\frac{a+i\cdot{b}\cos{\phi}}{c+i\cdot{d}\cos{\phi}}.$$
Using the Leibniz integral rule, take the derivative of $A$ wrt $\alpha$.
$$\frac{d}{d\alpha}A(\alpha) = \frac{d}{d\alpha} \int_{\alpha}^{2\pi+\alpha}\,d\phi\,\,\frac{a+i\cdot{b}\cos{\phi}}{c+i\cdot{d}\cos{\phi}}$$
$$ = \frac{a+i\cdot{b}\cos{(2\pi+\alpha)}}{c+i\cdot{d}\cos{(2\pi+\alpha)}} - \frac{a+i\cdot{b}\cos{\alpha}}{c+i\cdot{d}\cos{\alpha}}$$
$$ = \frac{a+i\cdot{b}\cos{\alpha}}{c+i\cdot{d}\cos{\alpha}} - \frac{a+i\cdot{b}\cos{\alpha}}{c+i\cdot{d}\cos{\alpha}}$$
$$ = 0 .$$
Hence, $A(\alpha)$ is a constant function.
