# Complex Laurent Series and Contour Integral

Let $$f(z) = \sin{(\frac{1}{z})}$$, where $$z \neq 0$$. Find a Laurent Series expansion of $$f$$ around the annulus $$D: 1< |z|<3$$.

Use the result to find $$\oint \limits_C z^4\sin{(\frac{1}{z})} dz$$ where $$C$$ is the curve described by $$|z|=2$$.

My attempt:

Since $$z=0$$ is the only singular point, but it is not contained in the annulus $$D$$, we have that $$f$$ is analytic inside the annulus. Hence, our Laurent Series will be precisely the Maclaurine Series of $$f$$ in $$D$$, that is, $$f(z) = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}\big( \frac{1}{z}\big)^{2n+1} ~~~ \text{where } z \in D$$

Now, for the second part of our question.

\begin{align}\oint \limits_C z^4\sin{(\frac{1}{z})} dz &= \oint \limits_C z^4 \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}\big( \frac{1}{z}\big)^{2n+1} dz \end{align}

This is the part where I am stuck and I an unsure how to continue. My trail of thought is the following, but I honestly do not know if this is correct:

Since, clearly $$z^4$$ is analytic everywhere and we know that our series is analytic inside our annulus $$D$$, the product of the two (our integrand), will be analytic everywhere in the intersection of their two domains of analyticity (I think I might just have made that word up, but you know what I mean :P ). Hence we have that our integrand is analytic in $$D$$.

Then, since $$C$$ is a closed, piecewise smooth curve inside $$D$$ and our integrand is analytic on $$C$$, we know, from Cauchy-Goursat, that $$\oint \limits_C z^4\sin{(\frac{1}{z})} dz =0$$

• Cauchy-Goursat assumes simply connected. You do not have that here. What happens when you integrate $\frac{1}{z}$ on a closed contour encircling the origin? What if you have any other power? Nov 12, 2015 at 6:34
• @CameronWilliams . Good point. Is there a way that I can make use of the Residue instead then, perhaps? Nov 12, 2015 at 6:38
• Yes residues are the way to go and this is exactly as I was suggesting in the original comment. Residues come from powers of $z^{-1}$. Nov 12, 2015 at 6:39
• @CameronWilliams . If our integrand is called $g$, then $\text{Res}[g, 0] = z^{4} |_{z=0} = 0$ , thus we know that $\oint_{C} g(z) = 2\pi i (0) =0$. Is this correct? :) Nov 12, 2015 at 6:47
• No. You have negative powers coming from $\sin(z^{-1})$. Write out the full Laurent series for $z^4\sin(z^{-1})$ term by term and you'll see what falls out. Nov 12, 2015 at 6:59

You are on the right track. The only non zero coefficient in the Laurent series is $z^{-1}$ or $n=2$. So $$\oint \limits_C z^4\sin{(\frac{1}{z})} dz = \oint \limits_C \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!} \frac{1}{z^{2n-3}} dz=\frac{1}{5!}\oint \frac{1}{z^{}} dz=\frac{1}{5!}\int_0^{2\pi} i d\theta=\frac{\pi i}{60}$$
• @mathlover - why do we disregard the minus of $-\frac{1}{5!}$? Nov 12, 2015 at 8:25