Please verify my classification of the following singularities

Question

There are 2 functions of interest.

1.) $z^2e^{1/z}$

2.)$\dfrac{z}{\cos z}$

My attempt:

For (1.) we have singularities at $0$ and $\infty$.

For $z = 0$, the singularity is essential because $\lim_{z\rightarrow0} z^2e^{1/z}$ does not exist either as $\infty$ or some finite complex number

For $z = \infty, $ the singularity is removable as $\lim_{z\rightarrow\infty} z^2e^{1/z}$exist.

For (2.)

We have singularities at $z = \infty$ and $z = \frac{\pi}{2} \ +\ n\pi$

For $z = \infty$, the limit is a pole since $\lim_{z\rightarrow\infty} \frac{z}{cosz}$ is $\infty$

For $z = \frac{\pi}{2} \ +\ n\pi$, I am having some problem classifying this one, so any help or insight on this is appreciated.

So basically above is my attempt, if I made any mistake please do point it out for me. (and if possible provide an explanation)

Answer 1

If you have an holomorphic function $f:U\longrightarrow\mathbb{C}$ with a zero in $z_0\in U$, then by using Taylor series, you can write in some open neighborhood of $z_0$, the function $f$ in the following form $$f(z)=(z-z_0)^nh(z),$$ where $n$ is the order of the zero $z_0$ and $h$ is holomorphic and non-zero in the chosen neighborhood. In particular note that $1/f(z)$ has a polus of order $n$ in $z_0$. In your case, the map $\cos z$ has a simple zero in $\pi/2$, since $\cos'(\pi/2)=\sin(\pi/2)\neq 0$, therefore $1/\cos z$ has a simple polus in $\pi/2$. By this last remark $z/\cos z$ has a simple polus in $\pi/2+n\pi$.

Answer 2

Arthur pointed out by asking what happens to the same points for $\frac{\cos(z)}{z}$ .

Recall the definition : "A function $f(z)$ has a singularity at $z=∞$ if $f(1/w)$ has a singularity at $w=0$ ." Since $\frac{\cos(z)}{z}$ has a singularity at $0$, what can you conclude about $\frac{z}{\cos(z)}$ ?