3
$\begingroup$

I have the following complex function: $$f(z)=\frac{\cos\left(\frac{\pi z}{2}\right)\sin\left(\frac{1}{z}\right)}{(z^2-1)(z-2)}$$

I kind of shown that $z=\pm 1$ and $z =2$ are simple poles. However I don't know how to show that $z=0$ is an essential singularity. How do I even find the Laurent series of $sin\left(\frac{1}{z}\right)$ around $0$?

I know started from the Taylor expansion $$\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^kz^{2k+1}}{(2k+1)!}$$ and then I wrote it for $\sin\left(\frac{1}{z}\right)$ $$\sin\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{(-1)^k}{(2k+1)!z^{2k+1}}$$ but then I don't know how to find the Laurent series of it, shouldn't it be from $-\infty$ to $\infty$? All I can do here seems to be $$\sin\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{(-1)^kz^{-(2k+1)}}{(2k+1)!}$$ which has negative exponents only?

Can you help me?

$\endgroup$
  • 2
    $\begingroup$ If $f(z) = \sum_{n=-\infty}^0 a_n z^n$ then we can also write $f(z) = \sum_{n=-\infty}^\infty a_n z^n$ where $a_1=a_2=a_3=\ldots = 0$. Thus if you only find negative exponents this just means that all the positive are zero. $\endgroup$ – Winther Dec 13 '16 at 23:29
  • $\begingroup$ @Winther, that makes sense. $\endgroup$ – Euler_Salter Dec 13 '16 at 23:30
2
$\begingroup$

$z = 0$ is an essential singularity because if you develop the function

$$\sin \frac{1}{z}$$

In powers of $z$, then ALL the terms show a pole at $z = 0$, and there is no way to remove them.

Hence it's indeed essential.

P.s. The Lurent series of $\sin$ is the actual Taylor series.

$\endgroup$
  • $\begingroup$ So what is the difference between a Laurent series and a Taylor series? Cause I thought the first was an extension of the second, including negative powers! But in this case there is no positive power at all, is there? $\endgroup$ – Euler_Salter Dec 13 '16 at 23:11
  • $\begingroup$ So in general how does someone say that a function has an essential singularity? Do you have to look at the Taylor/Laurent series and see where it is not defined? $\endgroup$ – Euler_Salter Dec 13 '16 at 23:12
  • $\begingroup$ @Euler_SalterWell the Taylor series only work when your function is holomorphic, the Laurent series works still for isolated singularities. When your $f(z)$ is holomorphic the Taylor series and the Laurent series are the same, and with Cauchy's theorem you can see that. If you want to be as good as possible you have to calculate those things on your own! ^^ $\endgroup$ – Von Neumann Dec 13 '16 at 23:14
  • $\begingroup$ @Euler_Salter Yes, that is the point. Laurent Series (or Taylor's) will tell you everything about a singularity. Take $f(z) = e^{1/z}$. Can you see why $z = 0$ is an essential singularity? $\endgroup$ – Von Neumann Dec 13 '16 at 23:14
  • $\begingroup$ well we should have $f(z) = e^{z^-1}= \sum_{n=0}^{\infty} \frac{(z^{-1})^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!z^n}$? Okay maybe I understand. So this is again an essential singularity because at every term we have that the element of the series is undefined at $z=0$? $\endgroup$ – Euler_Salter Dec 13 '16 at 23:17
2
$\begingroup$

Since $$ g(z)=\frac{\cos(\pi z/2)}{(z^2-1)(z-2)} $$ is holomorphic in a neighborhood of $0$ and $g(0)=1/2$, this factor can be removed for discussing the singularity at $0$ of $f$. Thus we remain with $$ h(z)=\sin\frac{1}{z} $$ which is certainly holomorphic in $0<|z|<1$, so $0$ is an isolated singularity. It is not removable, so it is either essential or a pole. If the latter, there should exist a positive integer $n$ such that $$ \lim_{z\to0}z^n\sin\frac{1}{z} $$ exists (finite) and is not $0$.

However, if we approach $0$ along the real axis, we have $$ \lim_{x\to0}x^n\sin\frac{1}{x}=0 $$ for every positive integer $n$. Hence the singularity is essential.


For the singularity at $1$, you should consider $$ \lim_{z\to1}\frac{\cos(\pi z/2)}{z-1}= \lim_{w\to0}\frac{\cos(\frac{\pi}{2}-w)}{-\frac{2w}{\pi}}= \lim_{w\to0}-\frac{\pi}{2}\frac{\sin w}{w}=-\frac{\pi}{2} $$ so $1$ is a removable singularity for $f$ and the same for $-1$.

If you want to see a different approach, let's look at $z=-1$. Granted that $$ \frac{1}{(z-1)(z-2)}\sin\frac{1}{z} $$ is holomorphic in a neighborhood of $-1$, we need to consider $$ g(z)=\frac{\cos(\pi z/2)}{z+1} $$ of which we want to find the Laurent series at $-1$; setting $z+1=w$, we have $$ \cos\left(\frac{\pi}{2}w+\frac{\pi}{2}\right)=-\sin\frac{\pi w}{2} $$ so $$ -\frac{1}{w}\sin\frac{\pi w}{2}= -\frac{1}{w}\left(\frac{\pi}{2}w- \frac{\pi^3}{2^3\cdot 3!}w^3+\dotsb\right)= -\frac{\pi}{2}+\frac{\pi^3}{48}w^2+\dotsb $$ so the singularity is removable.

$\endgroup$
  • $\begingroup$ first of all thank you for the long answer! Secondly, I have some questions: how did you deduce a priori that $z=0$ was not removable? $\endgroup$ – Euler_Salter Dec 13 '16 at 23:41
  • $\begingroup$ @Euler_Salter Since $\lim_{x\to0}\sin\frac{1}{x}$ doesn't exist even on the real line… $\endgroup$ – egreg Dec 13 '16 at 23:44
  • $\begingroup$ probably all these questions sound stupid, but I think I am not understanding the correlation between $\lim_{x\to 0} \sin\left(\frac{1}{z}\right)$ not existing and the singularity being not removable. The definition I found in the book of removable singularity is when there are no negative powers of $(z-a)$ which in this case is $z$ $\endgroup$ – Euler_Salter Dec 13 '16 at 23:49
  • $\begingroup$ @Euler_Salter The function $f$, holomorphic in $0<|z-a|<r$, has a removable singularity at $a$ if and only if $\lim_{z\to a}f(z)$ exists (finite). $\endgroup$ – egreg Dec 13 '16 at 23:56
  • $\begingroup$ okay I didn't know it. I know I am asking you a lot, but can you expand on why there exists $n$ etc.. ? Are you calling $n$ the order of the pole? $\endgroup$ – Euler_Salter Dec 14 '16 at 0:00

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.