For the function :

$$f(z)= \sin\bigg(\frac{1}{\cos(\frac1z)}\bigg)$$

the point $z=0$ is:

1)a removable singularity

2)a pole

3)an essential singularity

4)a non-isolated singularity

The answer seems to be 3)an essential singularity.

But I arrived at 1)removable singularity because when $f(z)$ has removable singularity,

$\lim\limits_{z\to0}$ $(z-z_0)$$f(z)=0$. (Since $\lim\limits_{x\to a}$$f(x)=f(a)$.)

Can someone help me, pointing out where I had gone wrong? Thanks in advance.

  • $\begingroup$ you may not plug in the value in a limit unless the function is continuous. The whole point is that the function is not continuous at 0. You are trying to determine how bad this discontinuity is $\endgroup$ – operatorerror Dec 29 '17 at 11:25

It has a non-isolated singularity at $0$, since it has a singularity at every point of the form $\frac1{\pi/2+n\pi}$ ($n\in\mathbb N$).

It is not true that $\lim_{z\to0}zf(z)=0$. Actually, this limit does not exist.

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  • $\begingroup$ I couldn't understand how is this a non-isolated singularity. Could you please explain again? I am convinced that it is not a removable singularity. We say, for example, $f(z) = cot z$ has isolated singularity at $z_0 = n\pi$. So how is this non-isolated? Thank you. $\endgroup$ – user517123 Dec 29 '17 at 12:22
  • $\begingroup$ @user517123 Because $0$ is a singularity and, as I proved, there are singularities arbitrarily close to $0$. That's what “non-isolated” mean. $\endgroup$ – José Carlos Santos Dec 29 '17 at 12:28
  • $\begingroup$ What about essential singularity? Doesn't have infinitely many negative power terms in the Laurent series expansion? $\endgroup$ – Unknown x Jun 14 at 2:12
  • $\begingroup$ @Unknownx The function $f$ has no Lauret expansion around $0$. $\endgroup$ – José Carlos Santos Jun 14 at 6:58
  • $\begingroup$ $\sin \left(\frac{1}{\cos \left(\frac{1}{z}\right)}\right)=\sec \left(\frac{1}{z}\right)-\frac{\sec ^3\left(\frac{1}{z}\right)}{3!}+\frac{\sec ^5\left(\frac{1}{z}\right)}{5!}-...\infty =\left(1+\frac{\left(\frac{1}{z}\right)^2}{2!}+\frac{5\left(\frac{1}{z}\right)^4}{4!}+...\right)-\frac{\left(1+\frac{\left(\frac{1}{z}\right)^2}{2!}+\frac{5\left(\frac{1}{z}\right)^4}{4!}+...\ \infty \right)^3}{3!}+...\infty $ $\endgroup$ – Unknown x Jun 14 at 13:48

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