Let $f(z)= \frac{2(e^\frac{1}{z}-1)(\sin^2z)}{z^3}$.
Calculate $\int\limits_{\partial B_+(O,1)} f(z)\operatorname{d}z$
Could someone confirm my solution?
Solution?
I try to calculate the residue at $z=0$.
Since $e^\frac{1}{z}=\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\ldots$ $$f(z)= \frac{1}{z^3}\left(-1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\ldots\right)\left(1-\cos 2z\right)$$ $$f(z)= \frac{1}{z^3}\left(-1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\ldots\right)\left(\frac{2^2z^2}{2!}-\frac{2^4z^4}{4!}+\frac{2^6z^6}{6!}-\ldots\right)$$
Since I would find the residue as the coefficient of $z^{-1}$, I'm looking for the terms containing $z^2$ originating from the last 2 factors.
This results in the following coefficients: $$-2-\frac{2^4}{2!4!}+\frac{2^6}{4!6!}-\frac{2^8}{6!8!}+\ldots = -2+\sum_{n=2}^{+\infty} \frac{(-1)^{n-1}2^{2n}}{(2n-2)!(2n)!}$$
As a conclusion using the residue theorem $$\int\limits_{\partial B_+(O,1)} f(z)\operatorname{d}z = -4\pi i +2\pi i\left( \sum_{n=2}^{+\infty} \frac{(-1)^{n-1}2^{2n}}{(2n-2)!(2n)!}\right)$$
I think it is correct, but the book where this problem originates from gives a solution without the term $-4\pi i$...