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I am having some trouble with the following question:

Show $$\int_C \frac{2+3\sin(\pi z)}{z(z-1)^2}\mathrm dz = -6\pi^2 i $$ with $C$ the circle of radius $6$, centre $0$, positively oriented.

I am assuming I use Cauchy's extended integration formula which states:

$$\int_C f(z) \mathrm dz = 2\pi i \sum_{k=1}^n \operatorname{Res}_{z=z_k} f(z)$$

However I am not sure what I should use as my $f(z)$.

If for one part $$f(z) = \frac{2+3\sin(\pi z)}{z} = \frac{1}{z} [2 + 3(1 - \frac{(\pi z)^3}{3!}+...)]$$ I have calculated already that the residue at $z = 0$ is equal to $5$.

I am not sure if my $f(z)$ should just be the original function though, or how I incorporate the $z = 1$ singularity.

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    $\begingroup$ Use partial fraction decomposition in the form $\frac{A}{z}+\frac{B}{(z-1)^2}+\frac{C}{z-1}$ and the cauchy-integral formula $\endgroup$ Jun 21, 2017 at 9:26
  • $\begingroup$ @Fakemistake I will attempt this but is it okay with the sine function? $\endgroup$
    – Inazuma
    Jun 21, 2017 at 9:36

1 Answer 1

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We have $$\frac{1}{z(z-1)^2}=\frac{1}{z}+\frac{1}{(z-1)^2}-\frac{1}{z-1}$$ Consequently for $f(z)=2+3\sin(\pi z)$ $$\oint_C\frac{f(z)}{z(z-1)^2}dz=\oint_C\Big(\frac{f(z)}{z}+\frac{f(z)}{(z-1)^2}-\frac{f(z)}{z-1}\Big)dz=2\pi i(f(0)+f'(1)-f(1))$$ Because of $f(0)=f(1)$ and $f'(1)=-3\pi$, the integral is equal to $-6\pi^2 i$


If you want to solve it with the residue theorem, you have to calculate the residues in the simple pole $z=0$ and double pole $z=1$. Define $$g(z)=\frac{2+3\sin(\pi z)}{z(z-1)^2}$$ Then $\mathrm{Res}_{z=0}g(z)=2$ and $\mathrm{Res}_{z=1}g(z)=-3\pi-2$.

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  • $\begingroup$ How do I split the sine function into partial fractions? $\endgroup$
    – Inazuma
    Jun 21, 2017 at 9:47

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