0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.