Singular points of $ \sin ( \frac{1}{\cos\frac{1}{z}})$

909 Views Asked by At

Specifically, $\displaystyle f(z) = \sin \left( \frac{1}{\displaystyle \cos \frac{1}{z}} \right)$ has singular points at $z = \displaystyle \frac{2}{\pi + 2\pi k}$, among others.

Now, I am trying to show that these are, in fact, essential singular points, but I am finding the process of doing so extremely difficult.

Writing out a Laurent Series would no doubt be extremely difficult, and so I was thinking about showing that the limit as $z \to \displaystyle \frac{2}{\pi + 2\pi k}$ does not exist, but that is also extremely hard.

Could somebody please help me get this problem done?

Thank you.

1

There are 1 best solutions below

8
On BEST ANSWER

If it's an isolated singularity, and neither a removable singularity nor a pole, then it must be an essential singularity. To show this it suffices to find a sequence $z_n$ approaching the singularity such that $f(z_n)$ does not have a finite limit and does not go to complex $\infty$. For example, $f(z_n)$ might alternate $\pm 1$.