How do I find singularities and residue? Given is $f(z)=\sin(\exp(\frac{1}{z}))$. How do I find singularities and residue?
I know that singularity for my function is $z_0=0$. But how do I find residue?
 A: The residue of $f$ at $0$ is
$$\frac{1}{2\pi i} \int_{\lvert z\rvert = r} f(z)\,dz,\tag{1}$$
where since $f$ is holomorphic on $\mathbb{C}\setminus \{0\}$ the radius $r$ can be chosen arbitrarily. (Generally, it must be chosen small enough that the circle doesn't enclose nor pass through any other singularity.)
Making the substitution $w = 1/z$ in $(1)$ yields - note that the substitution leads to a negatively oriented circle; we then use the minus sign of the derivative to change the orientation -
$$\operatorname{Res} (f;0) = \frac{1}{2\pi i} \int_{\lvert w\rvert = R} \frac{\sin (\exp w)}{w^2}\,dw$$
(where $R = 1/r$), which is easy to evaluate if one remembers the integral formula for derivatives.
A: As an alternative to the way forward presented by Daniel, we can expand the function $f(z)$ in a Laurent series.  Proceeding we write
$$\begin{align}
f(z)&=\sin\left(e^{1/z}\right)\\\\
&=\sin\left(1+\frac1z+O\left(\frac{1}{z^2}\right)\right)\\\\
&=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(1+\frac1z+O\left(\frac{1}{z^2}\right)\right)^{2n+1}\\\\
&=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}+\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\right)\,\frac1z+O\left(\frac{1}{z^2}\right)\\\\
\end{align}$$
The residue is, therefore, $\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}=\cos(1)$.
