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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 $$\oint \limits_C z^4\sin{(\frac{1}{z})} dz =0$$

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  • $\begingroup$ 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? $\endgroup$ Nov 12, 2015 at 6:34
  • $\begingroup$ @CameronWilliams . Good point. Is there a way that I can make use of the Residue instead then, perhaps? $\endgroup$
    – user860374
    Nov 12, 2015 at 6:38
  • $\begingroup$ Yes residues are the way to go and this is exactly as I was suggesting in the original comment. Residues come from powers of $z^{-1}$. $\endgroup$ Nov 12, 2015 at 6:39
  • $\begingroup$ @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? :) $\endgroup$
    – user860374
    Nov 12, 2015 at 6:47
  • $\begingroup$ 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. $\endgroup$ Nov 12, 2015 at 6:59

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

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  • $\begingroup$ @mathlover - why do we disregard the minus of $-\frac{1}{5!}$? $\endgroup$
    – user860374
    Nov 12, 2015 at 8:25
  • $\begingroup$ Thank you :). That makes perfect sense :) $\endgroup$
    – user860374
    Nov 12, 2015 at 8:47

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