Residue calculation with complex conjugate denominator Let $a_1, a_2, \lambda_1, \lambda_2 \in \mathbb{C}$ constants and define 
$$f: s \mapsto \sum_{i=1}^2 \dfrac{a_i}{s-\lambda_i}$$
I need to calculate the residue with respect to $\lambda_1$ and $\lambda_2$ of the function $\overline{f(s)} f(s)$.
The question is how to this (the complex conjugate term frustrates me). I don't know much anymore about calculating residues so I have following question.
\begin{eqnarray}
Res( \overline{f(s)} f(s), \lambda_1) & = Res( \dfrac{ \overline{a_1} a_2}{(\overline{s}-\overline{\lambda_1})(s-\lambda_2)}, \lambda_1) + Res( \dfrac{ \overline{a_1} a_1}{(\overline{s}-\overline{\lambda_1})(s-\lambda_1)}, \lambda_1) + \\
&  Res( \dfrac{ a_1 \overline{a_2}}{(\overline{s}-\overline{\lambda_2})(s-\lambda_1)}, \lambda_1) \\
& = Res( \dfrac{ \overline{a_1} a_2}{(\overline{s}-\overline{\lambda_1})(s-\lambda_2)}, \lambda_1) + Res( \dfrac{ \overline{a_1} a_1}{(\overline{s}-\overline{\lambda_1})(s-\lambda_1)}, \lambda_1) + a_1 \overline{a_2}\\
\end{eqnarray}
I don't know how to deal with the complex conjugates. Can anyone help me?
 A: We know that $\int_{C}z^ndz=2\pi[n=-1]$, where $C$ is any small circle around the origin and $[n=-1]=1$ when $n=-1$ and zero otherwise.
Similarly, we can compute $$\int_Cz^m\overline{z}^ndz=\int_Cz^{m-n}dz=2pi[m-n=-1]$$
In the equalities above we scaled first the circle to the unit circle and then used that for $|z|=1$ we have $\overline{z}=z^{-1}$.
We can use this to compute all those residues.

Let me just do one of them.
\begin{align}
\frac{\overline{a_1}a_2}{\overline{(s-\lambda_1)}(s-\lambda_2)}&=\frac{\overline{a_1}a_2}{\overline{(s-\lambda_1)}(s-\lambda_1+\lambda_1\lambda_2)}\\&=\frac{\overline{a_1}a_2}{(\lambda_1-\lambda_2)}\frac{1}{1+\frac{s-\lambda_1}{\lambda_1-\lambda_2}}\overline{(s-\lambda_1)}^{\ -1}\\&=\frac{\overline{a_1}a_2}{(\lambda_1-\lambda_2)}\sum_{m=0}^{\infty}(-1)^{m}\frac{(s-\lambda_1)^m\overline{(s-\lambda_1)}^{\ -1}}{(\lambda_1-\lambda_2)^m}
\end{align}
Using our computations the integrals of all the terms are going to be zero except when $m-(-1)=-1$. There is no $m=-2$ in the sum, therefore the residue is zero for $\frac{\overline{a_1}a_2}{\overline{(s-\lambda_1)}(s-\lambda_2)}$ at $s=\lambda_1$.

Well, let me compute one that you seem to have computed in the question.
\begin{align}
\frac{a_1\overline{a_2}}{(\overline{s}-\overline{\lambda_2})(s-\lambda_1)}&=\frac{a_1\overline{a_2}}{(\overline{s}-\overline{\lambda_2})(s-\lambda_1)}\\&=\frac{a_1\overline{a_2}}{(\overline{s}-\overline{\lambda_1}+\overline{\lambda_1}-\overline{\lambda_2})(s-\lambda_1)}\\&=\frac{a_1\overline{a}_2}{\overline{(\lambda_1-\lambda_2)}}\frac{1}{1+\frac{\overline{s-\lambda_1}}{\overline{(\lambda_1-\lambda_2)}}}\frac{1}{s-\lambda_1}\\&=\frac{a_1\overline{a}_2}{\overline{(\lambda_1-\lambda_2)}}\sum_{n=0}^{\infty}(-1)^n\frac{\overline{(s-\lambda_1)}^{n}(s-\lambda_1)^{-1}}{\overline{(\lambda_1-\lambda_2)}^n}
\end{align}
Using our computations above the integral around $s=\lambda_1$ of all those terms are zero except when -1+n=-1. This means that for $n=0$ we get some non-zero residue. We get $2\pi\frac{a_1\overline{a_2}}{\overline{\lambda_1-\lambda_2}}$.
