find all the singular points and residues of given function

Question

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?

Answer 1

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 $f$?

Answer 2

The residue at $z=-2$ is straightforward to determine by

$$\text{Res}\left(\frac{e^{1/z}}{(z+2)^2}, z=-2\right)=\lim_{z\to -2}\frac{d}{dz}\left((z+2)^2\frac{e^{1/z}}{(z+2)^2}\right)=-\frac14 e^{-1/2}$$

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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

$$\text{Res}\left(\frac{e^{1/z}}{(z+2)^2}, z=0\right)=\frac14\sum_{m=0}^\infty \frac{m+1}{(m+1)!\,2^m}=\frac14 e^{-1/2}$$

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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!