I would like to check and see if my reasoning for this question is correct:
Find the singular points of the function, and classify them if they are isolated singular points. Also, evaluate if $\infty$ is an isolated singular point.
- $\cot \frac{1}{z}$
My reasoning for this question is similar to that found in this link below:
Find and classify singular points of $\cot\left(\frac{1}{z}\right)$
- $e^{-z} \cos \frac{1}{z}$
$z = 0$ IS an isolated singular point this time, because $e^{-z}$ is entire, making the function analytic everywhere else. This is the only statement I'd like to check.
- $e^{\cot(\frac{1}{z})}$
- $\cot \frac{1}{z} - \frac{1}{z}$
Would we apply the same reasoning as that found in the link I posted above? That is, since $z = 0$ is not an isolated singular point of $\cot \frac{1}{z}$, the same is true the functions above?
- $\sin{(\frac{1}{\cos {\frac{1}{z}}})}$
Not sure where to start with this one, other than perhaps identifying that $\cos \frac{1}{z}$ has multiple zeroes, making our function non-analytic in multiple points. Therefore, since the function is non-analytic for several points for all neighborhoods around $z = 0$, this is also a non-isolated singular point.
Please let me know if my logic is right, or if there's something else that needs to be addressed. Thank you!