Determine all singularities and its characters for function: $f(z) = \frac{\sin (z+1) e^{\frac{1}{z}}}{(z-i)^{2}(z+i)\cos ^{2} (z)}.$

Question

Determine all singularities and its characters for function: $$f(z) = \frac{\sin (z+1) e^{\frac{1}{z}}}{(z-i)^{2}(z+i)\cos ^{2} (z)}.$$

I have concluded that $z=0$ is essential singularity, $z=i$ order $2$ pole, $z=-i$ order $1$ pole.

1.question : Is that correct? 2.question : Can you help me with all $z =$ +/- $i\frac{\pi}{2} + 2k\pi, k\in \mathbb{Z}$ ? At first I thought of developing $f$ in Laurent's series, but I got stuck.

Answer 1

Let's write down all points that can be singularities for $f(z)$:

1. $z = i$

2. $z = -i$

3. $z = 0$

4. $\cos^2 (z) = 0$

5. $z = \infty$

As you mentioned, 1 - order 2 pole, 2 - order 1 pole, 3 = is essential singularity. I think, you can easily proof it.

Let's talk about 4. From $\cos^2 (z) = 0$ we get: $z_n = \pi/2 + \pi n$

Function $g(z) = \frac{\sin (z+1) e^{\frac{1}{z}}}{(z-i)^{2}(z+i)}$ is holomorphic in $z_n$, that's why $z_n$ is pole. Insofar as first derivative of $\cos(z)^2$ is $-2\cos(z)\sin(z)$ and equal to $0$ in $z_n$, but not the second derivative, we get that $z_n$ is order 2 pole of $f(z)$.

Thus, $\infty$ is not singularity, because it's a limit of singularities.