Singularities. Residues

Question

Classify the singularities and calculate the residual of each of them of the function $$f(z) = \frac{1}{z-1} e^{\frac{1}{(z-1)^2}} + \frac{z}{sin(z)}$$

Answer 1

First, I denote $$\left\lbrace \begin{array}{l} g(z) = \frac{1}{z-1} e^{\frac{1}{(z-1)^2}} \\ h(z) = \frac{z}{sin(z)} \end{array} \right.$$

Next, I note that $$\left\lbrace \begin{array}{l} g\in \mathcal{H}(\mathbb{C} \setminus \{1\}) \\ h\in \mathcal{H}(\mathbb{C} \setminus \{k\pi : k\in \mathbb{Z}\}) \end{array} \right.$$

So, $f\in \mathcal{H}(\mathbb{C} \setminus \{1, k\pi : k\in \mathbb{Z}\})$.

Singularities

1. Case $z=1$

   • $h$ has not singularities in $z=1$
   • $g$ has a essential singularity in $z=1$ because $$g(z) = \sum_{n=-\infty}^{0} b_n (z-1)^n$$ where $$b_n = \left\lbrace \begin{array}{l} \frac{1}{(-k)!} \text{ if } n=2k-1 \\ 0 \text{ other case} \end{array} \right.$$ I.e. $b_{n}$ has infinity terms that are not zero.
   • Total, $f$ has essential singularity in $z=1$

2. Case $z=0$

   • $g$ has not singularities in $z=0$
   • $h$ has a removable singularity in $z=0$ because $$\lim_{z \rightarrow 0} \frac{z}{\sin(z)} = 1 $$
   • Total, $f$ has removable singularity in $z=0$

3. Case $z=k\pi : k\in\mathbb{Z}, k\not=0$

   • $g$ has not singularities in $z=k\pi$
   • $h$ has a pole in $z=k\pi$ because $$\lim_{z \rightarrow k\pi} \frac{z}{\sin(z)} = \infty$$ Its order is -1

Residues

1. Case $z=1$.

   • I need calculate the Laurent series of $f(z)$ in $z=1$.
   • Since $h \in \mathcal{H}(B(1,1))$, $$g = \sum_{n=0}^{\infty} a_n (z-1)^n$$ where $a_n$ is determinated by unicity.
   • Since I calculated before the expression of $h$ in Laurent Series, we have $$g(z) = \sum_{n=-\infty}^{0} b_n (z-1)^n$$ where $$b_n = \left\lbrace \begin{array}{l} \frac{1}{(-k)!} \text{ if } n=2k-1 \\ 0 \text{ other case} \end{array} \right.$$
   • Total $$\text{Res}(f, z=1)= \underbrace{\text{Coef}_g \left(\frac{1}{z-1}\right)}_{1/0!} + \underbrace{\text{Coef}_h \left(\frac{1}{z-1}\right)}_{0} = 1 $$

2. Case $z=0$

   • Since $z=0$ is a removable singularity of $f$, $$\text{Res}(f, z=0) = 0$$

3. Case $z=k\pi$, $k\in \mathcal{Z}$, $k\not=0$

   • Since $z=k\pi$ is a pole of order -1, $$\text{Res}(f, z=k\pi) = \lim_{z\rightarrow k\pi} (z-k\pi) f(z) = 0 + \frac{2k\pi}{\cos(k\pi)} $$