Let $a,b \in \mathbb{C}$, with $|b|<1$. Calculate the integral $$\frac{1}{2\pi i}\int_{|z|=1}\frac{|z-a|^2}{|z-b|^2}\frac{dz}{z}. $$
I know that $$\frac{1}{2\pi i}\int_{|z|=1}\frac{|z-a|^2}{(z-b)(\bar{z}-\bar{b})}\frac{dz}{z}. $$ This has a residue of order 1 at $z=b$ and at $z=0$.
Question 1: My professor said that because $z\bar{z}=1$, $\frac{1}{z}=\bar{b}$ so $z=\frac{1}{\bar{b}}$, but doesn't this restrict our solutions to the boundary? Wouldn't we want to know what makes $\bar{z}-\bar{b}=0$ no matter if $z\bar{z}=1$ or no? Why do we make this assumption?
Question 2: Okay, say fine, I accept the above and continue with the problem. then, I can assume that $\frac{1}{\bar{b}}>1$ and thus outside of our contour.
For a singularity of order $n$: Res{$f,z_0$} $= \frac{1}{(n-1)!}\lim_{z\rightarrow z_0}\frac{d^n}{dz^n} \big((z-z_0)^n f(z)\big)$. I also know that if $\phi(z_0)\neq 0$ and $z_0$ is a singularity of order $n$ then the Res{$\frac{\phi(z)}{g(z)},z_0$} $=$ $\frac{\phi(z_0)}{g'(z_0)}$
Res{$f,b$} $=\lim_{z\rightarrow b}\frac{|z-a|^2}{z(\bar{z}-\bar{b})}$ but then doesn't this diverge? If I use that handy trick (again, not sure why we can use this, isn't it restricting our answer?) then Res{$f,b$} $=\lim_{z\rightarrow b}\frac{|z-a|^2}{z(z-\frac{1}{\bar{b}})}=\frac{|b-a|^2}{b(b-\frac{1}{\bar{b}})}$ .
Res{$f,0$} $=\frac{|z-a|^2}{|z-b|^2}=\frac{|a|^2}{|b|^2}$
$$\frac{1}{2\pi i}\int_{|z|=1}\frac{|z-a|^2}{|z-b|^2}\frac{dz}{z} = \text{Res}\{f,b\}+\text{Res}\{f,0\} =\frac{|a|^2}{|b|^2}+\frac{|b-a|^2}{b(b-\frac{1}{\bar{b}})}.$$
Do I have to do a special case for $b=0?$ $$\frac{1}{2\pi i}\int_{|z|=1}\frac{|z-a|^2}{|z|^2}\frac{dz}{z} =\frac{1}{2\pi i}\int_{|z|=1}\frac{|z-a|^2}{z^2\bar{z}}dz$$
Pole of order 3?
$$\text{Res}\{f,0\} = \lim_{z\rightarrow 0 } \frac{d^2}{dz^2} z\frac{|z-a|^2}{\bar{z}} $$ and now I am stuck.
On another note, I know
$$\frac{n!}{2\pi i}\int_C \frac{f(z)}{(z-a)^{n+1}}dz=f^{(n)}(a) \text{ and } \frac{1}{2\pi i}\int_C\frac{f(z)}{(z-a)}dz=f(a) $$
So I would have (this has got to be wrong because I am disregarding the absolute values sign) $$f^1(b)=\frac{1!}{2\pi i}\int_C\frac{\frac{|z-a|^2}{z}}{|z-b|^2} =2\pi i \frac{|b-a|^2}{b}$$ and $$f(0)=\frac{1}{2\pi i } \int_C \frac{\frac{|z-a|^2}{|z-b|^2}}{z}dz=\frac{2\pi i |a|^2 }{|b|^2} $$
You are integrating $\frac {(z-a)(\frac 1 z - \overline {a})} {(z-b)(\frac 1 z - \overline {b})z}$. Poles inside the contour are at $0$ and $b$. the residues are $\frac a b$ and $\frac {(b-a) (1-\overline {a} b)} {b(1-|b|^{2})}$. So the answer is $\frac 1 b + \frac {(b-a) (1-\overline {a} b)} {b(1-|b|^{2})}$.