Let $a,b \in \mathbb{C}$ with $|b| < 1$. Compute $$\frac{1}{2\pi i} \oint_{|z|=1} \frac{|z-a|^2}{|z-b|^2} \frac{dz}{z}.\tag{1}$$
I'm studying this problem from a past QR exam for my upcoming exam. I think I have a correct solution, but I think I'm not understanding what's going on with integrating a non-analytic function.
Here's my solution: The main trick here is that since we're integrating $z \in \{|z|=1\}$, then $\overline{z} = 1/z$, and so
\begin{align} \frac{1}{2\pi i} \oint_{|z|=1} \frac{(z - a)(\overline{z} - \overline{a})}{(z - b)(\overline{z} - \overline{b})} \frac{dz}{z} &= \frac{1}{2\pi i} \oint_{|z|=1} \frac{(z - a)\left(\frac{1}{z} - \overline{a}\right)}{(z-b)\left(\frac{1}{z} - \overline{b}\right)} \frac{dz}{z} \\ &= \frac{1}{2\pi i} \oint_{|z|=1} \frac{(z - a)(1 - \overline{a}z)}{z(z-b)(1 - \overline{b}z)} \,dz.\tag{2} \end{align}
Because
$$|b| < 1 \Longleftrightarrow \frac{1}{|b|} = \frac{1}{|\overline{b}|} > 1,$$
then, we need only to find the residues at $z = 0, b$. Letting $f$ be the integrand in (2),
\begin{align} \frac{1}{2\pi i} \oint_{|z|=1} \frac{|z-a|^2}{|z-b|^2} \frac{dz}{z} &= \text{Res}(f;z=0) + \text{Res}(f;z=b) \\ &= \frac{a}{b} + \frac{(b - a)(1 - \overline{a}b)}{b(1 - |b|^2)} \\ &= \frac{|a|^2 - 2\text{Re}(\overline{a}b) + 1}{1 - |b|^2}. \end{align}
Now, assuming this is correct, what confuses me a little here is that $1/\overline{b}$ is not really a pole of $|z-a|^2/|z-b|^2$, but it is when you integrate along the unit circle? If the contour were $|z| = 2$ and, say, $|b| < 1/2$. Then, the residues will be different, even though the poles are the same. (The pole at $2/\overline{b}$ is outside $|z|=2$.) Is this explained by the behavior of $|z - a|^2/|z-b|^2$ not being analytic anywhere (except maybe at $a$, not sure)?