# Computing a complex integral using residues.

While revising for my complex analysis exam in a few weeks, I have come across a tricky problem. I haven't done much work around the residue theorem yet and so I can't solve many examples. Here is the question:

Compute the integral $\int_{C_r(0)}$sin$(\frac{1}{z})dz$ using residues where $r >0$ and $\int_{C_r(w)}f(z)dz$ = $\int_{\gamma,r|[0,2π]}f(z)dz$.

I have tried to use the residue theorem but I can't seem to calculate the residues correctly. Any help is appreciated.

## 2 Answers

Since $\displaystyle\sin\left(\frac1z\right)=\frac1z-\frac1{3!z^3}+\frac1{5!z^5}-\cdots$, $\operatorname{res}_0\left(\sin\left(\frac1z\right)\right)=1$.

• I got that expansion but am unsure of how you got the residue at 0. Is there a theorem I'm missing? – Zombiegit123 May 4 '18 at 15:35
• @Zombiegit123 No, there's a definition that you're missing: the residue at $0$ is by definition the coefficient of $\frac1z$. – José Carlos Santos May 4 '18 at 15:36
• Ok I see, how do I go from here? – Zombiegit123 May 4 '18 at 15:51
• @Zombiegit123 $\displaystyle\int_{C_r(0)}\sin\left(\frac1z\right)\,\mathrm dz=2\pi i\operatorname{res}_0\left(\sin\left(\frac1z\right)\right)=2\pi i.$ – José Carlos Santos May 4 '18 at 15:53
• i didnt realise it was this simple. I thought you had to sum all of the residues? Thanks for the help. – Zombiegit123 May 4 '18 at 15:55

Hint. Note that the Laurent expansion of $\sin(1/z)$ at $0$ is $$\frac{1}{z}+\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)! z^{2k+1}}.$$