find all the singular points and residues of given function I have a function:
$$f(z) = \frac{e^{\frac{1}{z}}}{(z+2)^2}$$
do I understand correctly that I have to find singular points in the numerator as well as denumerator? Also how am I supossed to find residues of the given function if it has different singularity types? Should I rewrite it like this?
$$f(z) = e^{\frac{1}{z}}\cdot\frac{1}{(z+2)^2}$$
and therefore $e^{\frac{1}{z}}$ can be treated as especial singularity and be expanded to the laurent series with $\frac{1}{z} = t$ and see then which summand triggers $z^{-1}$ after expansion?
and $\frac{1}{(z+2)^2}$ should be treated as a part that has second-order pole and therefore standard formula should be applied here?
 A: The function is not defined at $0$ and $-2$; it is otherwise holomorphic in a neighborhood of all other points.
Since $e^{1/z}$ is holomorphic in a neighborhood of $-2$, we clearly have that $-2$ is a pole of order $2$ for $f$
Since $(z+2)^{-2}$ is holomorphic in a neighborhood of $0$ and $e^{1/z}$ has an essential singularity at $0$ (your justification is good), $0$ is also an essential singularity for $f$.
In order to find the residue at $-2$, it's better to consider $w=z+2$ and find the residue of $g(w)=w^{-2}e^{1/(w-2)}$ at $0$. If
$$
e^{1/(w-2)}=\sum_{k\ge0}a_nw^k
$$
then
$$
g(w)=\sum_{k\ge-2}a_{k+2}w^k
$$
and therefore the residue is $a_1$; can you compute it?
For the residue at $0$, consider $u=2/z$, and
$$
h(u)=\frac{1}{4}e^{u/2}\frac{u^2}{(1+u)^2}
$$
which has a zero of order $2$ at $0$, so it can be written as
$$
\sum_{k\ge2}b_ku^k
$$
What can you derive from this about the Laurent series for $f$?
A: The residue at $z=-2$ is straightforward to determine by
$$\text{Res}\left(\frac{e^{1/z}}{(z+2)^2}, z=-2\right)=\lim_{z\to -2}\frac{d}{dz}\left((z+2)^2\frac{e^{1/z}}{(z+2)^2}\right)=-\frac14 e^{-1/2}$$

DIRECT COMPUTATION OF THE RESIDUE AT THE ORIGIN
To calculate the residue at $z=0$, we proceed as follows.  First, the function $e^{1/z}$ has Laurent series representation 
$$e^{1/z}=\sum_{n=0}^\infty \frac{z^{-n}}{n!}\tag 1$$
Second, the series representation for $\frac{1}{(z+2)^2}$ is 
$$\begin{align}
\frac{1}{(z+2)^2}&=\frac{1}{4\left(1+\frac z2\right)^2}\\\\
&=\frac14\sum_{n=0}^\infty (-1)^n(n+1)(z/2)^n\tag3
\end{align}$$
for $|z|<2$.
Third, using $(1)$ and $(2)$, the Cauchy Product for the product $e^{1/z}\times \frac1{(z+2)^2}$ is given by 
$$\sum_{n=0}^\infty \sum_{p=0}^n \frac{z^{-(n-p)}}{(n-p)!}\frac14 (-1)^p (p+1)(z/2)^p\tag 3$$
The residue of the product $e^{1/z}\times \frac1{(z+2)^2}$ is the coefficient on $z^{-1}$, which occurs when $p=(n-1)/2$ in $(1)$.  This only occurs when $n$ is odd.  Hence, we find from $(3)$ that
$$\text{Res}\left(\frac{e^{1/z}}{(z+2)^2}, z=0\right)=\frac14\sum_{m=0}^\infty \frac{m+1}{(m+1)!\,2^m}=\frac14 e^{-1/2}$$

INDIRECT COMPUTATION OF THE RESIDUE AT THE ORIGIN
An alternative approach to evaluating the residue at $z=0$ is to note that 
$$\begin{align}
\lim_{R\to \infty }\oint_{|z|=R}\frac{e^{1/z}}{(z+2)^2}\,dz&=\lim_{R\to \infty }\int_0^{2\pi}\frac{e^{1/(Re^{i\phi})}}{(Re^{i\phi}+2)^2}\,iRe^{i\phi}\,d\phi\\\\
&=0\\\\
&=2\pi i \left(\text{Res}\left(\frac{e^{1/z}}{(z+2)^2}, z=-2\right)+\text{Res}\left(\frac{e^{1/z}}{(z+2)^2}, z=0\right) \right) 
\end{align}$$
Hence the residues at $z=0$ and $z=-2$ are of opposite sign.  And given the residue at $z=-2$ is $-\frac14 e^{-1/2}$, we find that the reside at $z=0$ is $\frac14 e^{-1/2}$ as obtained by direct computation!
