I cannot seem to get my head around a part from Stephen Leduc's Cracking the GRE Mathematics Subject Test:
Let's figure out $$ \oint_{C}\frac{4z}{(z-1)(z-3)^2}dz $$ where $C$ is the circle $|z|=2$, oriented counterclockwise. The Laurent series for the integrand is: $$ f(z)=\frac{4z}{(z-1)(z-3)^2}=\sum^\infty_{n=1}z^{-n}+\sum^\infty_{n=0}\frac{2n+3}{3^{n+1}}z^n\text{, for }1<|z|<3 $$ which is valid in an annulus that contains the circle $C$. If we expand this Laurent series as a polynomial, the coefficient of the $z^{-1}$ term in this series is $a_{-1}=1$, so $$ \oint_{C}\frac{4z}{(z-1)(z-3)^2}dz=2\pi i\cdot a_{-1}=2\pi i $$
I have a few questions here:
Why can we use the Laurent series around $0$ if the pole is actually at $x=1$?
Why is $Res(0,f)\neq 0$ if $f$ is analytic at $0$ (i.e., $0$ is not a singularity)?
Ultimately, why is the answer still agree with which done by the simple pole method, $Res(1,f)=\lim_{z\to1}(z-1)f(z)=1$?