Singularities of $f(z)=z/\cos(z)$ Regarding complex functions (in complex variables), I was wondering why the function $g(z)= \cos(z)$ has a singularity at $z = \infty$ but $f(z)= \dfrac{z}{\cos(z)}$ does not.

I am a bit confused about the concept of singularities at $\infty$.

Thank you.
So, to sum up after all these comments:

$g(z)$ indeed has an essential singularity at $z=\infty$ because the function $h(z)=g\big(\frac1z\big)$ has an essential singularity at $z=0$:
$$h(z) = \cos\left(\frac1z\right) = 1- \frac1{2!}\left(\frac1z\right)^2 + \frac1{4!}\left(\frac1z\right)^4+ \dots $$
has infinitely many terms in its Laurent expansion at $0$.

On the other hand, $f(z)$, all appearances to the contrary, does not have an essential singularity at $\infty$: $f(z)$ has poles at $z=z_n=n\pi + \pi/2$ and $z_n\to\infty$, and so $\infty$ is not an isolated singular point.
For $\,\frac1z\,$ "close enough to zero" ( say, $\;\frac1{|z|}&lt;\sqrt[3]6\;$ ) , we have

$$f\left(\frac1z\right)=\frac1{z\cos\frac1z}=\frac1z\frac1{\left(1-\frac1{2z^2}+\frac1{24z^4}-\ldots\right)}=$$

$$=\frac1z\left(1+\frac1{2z^2}+\frac1{4z^4}+\ldots\right)=\frac1z+\frac1{2z^3}+\ldots$$

which shows $\,f(z)\;$ has a singularity at $\,z=\infty\;$