# Using Residue Theorem with a Laurent series not around a pole

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, $Res(1,f)=\lim_{z\to1}(z-1)f(z)=1$?

• $Res(0,f)$ is $0$. Residue's theorem tells you that the integral of $f(z)$ over a closed curve $C$ is equal to the sum of the residues (given there is a finite number of them) of $f$ in the interior of the curve $C$ times $2\pi i$. Here the only singularity of $f$ in the interior of $C$ is at $z_0 = 1$ so you only consider $Res(1,f)$ to compute your integral. – Desura Dec 29 '16 at 19:05
• @Desura My first bulletpoint is that the residue at $z_0=1$ is supposed to be the coefficient of $(z-z_0)^{-1}$, instead of $z^{-1}$. The way the book writes makes me understood as the latter way. – underlandian Dec 29 '16 at 23:42
• Yep, you're right. It should be $(z-1)^{-1}$, $\lvert z \rvert \lt 1$. Maybe they did a mistake or they did some kind of change of variables, but the equality would still be wrong as written in this state, – Desura Dec 29 '16 at 23:52
• @Desura what they argued was that all it needs was the path being in the annulus, so that is where it got me confused. – underlandian Dec 31 '16 at 4:13

Note that because 3 isn't into the region If we define $g(z)=\frac{4z}{(z-3)^2}$, by Cauchy's formula, this integral is equal to $2\pi i g(1)=2\pi i$.