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

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  • $\begingroup$ $Res(0,f)$ 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. $\endgroup$ – Desura Dec 29 '16 at 19:05
  • $\begingroup$ @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. $\endgroup$ – underlandian Dec 29 '16 at 23:42
  • $\begingroup$ 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, $\endgroup$ – Desura Dec 29 '16 at 23:52
  • $\begingroup$ @Desura what they argued was that all it needs was the path being in the annulus, so that is where it got me confused. $\endgroup$ – underlandian Dec 31 '16 at 4:13
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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 $2\pi i g(1)=2\pi i$.

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  • $\begingroup$ Well that is the simple pole method -- that I understand. I am just trying to understand the way the book does. $\endgroup$ – underlandian Dec 29 '16 at 23:35

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