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 😛 ). 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 $$\oint \limits_C z^4\sin{(\frac{1}{z})} dz =0$$
Best Answer
You are on the right track. The only non zero coefficient in the Laurent series is $z^{-1}$ or $n=2$. So $$ \oint \limits_C z^4\sin{(\frac{1}{z})} dz = \oint \limits_C \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!} \frac{1}{z^{2n-3}} dz=\frac{1}{5!}\oint \frac{1}{z^{}} dz=\frac{1}{5!}\int_0^{2\pi} i d\theta=\frac{\pi i}{60} $$