1. ## [SOLVED] Pole/essential singularity

I must find the singularities and classify them (poles, essential or removable singularities) and find their principal part.
The function is $f(z)=\frac{\sin z}{z^2(z-\pi)}$.

My attempt :
By a first look, I notice that the function has a pole or singularity at $z=0$ and $z=\pi$.
Let's examinate what happens when $z=0$.
The Taylor series of $\sin z$ in $z=0$ is $\sum _{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}$.
Thus $f(z)=\frac{1}{z(z-\pi)}-\frac{z}{z-\pi}+\frac{z^3}{z-\pi}-\frac{z^5}{z-\pi}+...$.
Now I'm unsure of myself. I see that the first term is undetermined when $z=0$ and $z=\pi$ while all the other terms are undetermined only when $z=\pi$. As a consequence I'm tempted to say that $z=0$ is a pole of order $1$. (Isn't a removable singularity also?).

$\pi$ would be an essential singularity since there are infinite terms where $z=\pi$ is not defined, but I don't know how to justify formally.

Plus, another doubt I have is why did I find the Taylor series of $f$ in $z=0$? I could have done it for an arbitrary $z_0$, or not?

2. Originally Posted by arbolis
I must find the singularities and classify them (poles, essential or removable singularities) and find their principal part.
The function is $f(z)=\frac{\sin z}{z^2(z-\pi)}$.

My attempt :
By a first look, I notice that the function has a pole or singularity at $z=0$ and $z=\pi$.
Let's examinate what happens when $z=0$.
The Taylor series of $\sin z$ in $z=0$ is $\sum _{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}$.
Thus $f(z)=\frac{1}{z(z-\pi)}-\frac{z}{z-\pi}+\frac{z^3}{z-\pi}-\frac{z^5}{z-\pi}+...$.
Now I'm unsure of myself. I see that the first term is undetermined when $z=0$ and $z=\pi$ while all the other terms are undetermined only when $z=\pi$. As a consequence I'm tempted to say that $z=0$ is a pole of order $1$. (Isn't a removable singularity also?).

$\pi$ would be an essential singularity since there are infinite terms where $z=\pi$ is not defined, but I don't know how to justify formally.

Plus, another doubt I have is why did I find the Taylor series of $f$ in $z=0$? I could have done it for an arbitrary $z_0$, or not?
$\lim_{z \rightarrow \pi} f(z) = - \frac{1}{\pi^2}$ using l'Hopital's Rule so $z = \pi$ is clearly a removable singularity.

Since $\lim_{z \rightarrow 0} z f(z) = - \frac{1}{\pi}$ it can be said that $z = 0$ is a simple pole. You should calculate the Laurent series for $f(z)$ in powers of z and confirm that the principle part is $- \frac{1}{\pi} \frac{1}{z}$.

3. Thanks for the help!

I have a question, you wrote
Since it can be said that is a simple pole.
, but why did you multiplied $f(z)$ by $z$? (I understand the rest assuming that I understand the reason of the product)

4. Originally Posted by arbolis
Thanks for the help!

I have a question, you wrote , but why did you multiplied $f(z)$ by $z$? (I understand the rest assuming that I understand the reason of the product)
Think about what the principle part of the Laurent series is for a function $g(z)$ that has a pole of order n at $z = z_0$. What must happen if $g(z)$ is then multiplied by $(z - z_0)^n$ and what does that mean about $\lim_{z \rightarrow z_0} (z - z_0)^n g(z)$ ....?

5. Originally Posted by arbolis
Thanks for the help!

I have a question, you wrote , but why did you multiplied $f(z)$ by $z$? (I understand the rest assuming that I understand the reason of the product)

This is a definition: the point $a$ is a simple pole of a function $f(z)$ if there exists an analytic function $g(z)$ s.t $f(z)=\frac{g(z)}{z-a}\, \forall{z\neq{a}}$

Tonio

6. Originally Posted by mr fantastic
Think about what the principle part of the Laurent series is for a function $g(z)$ that has a pole of order n at $z = z_0$. What must happen if $g(z)$ is then multiplied by $(z - z_0)^n$ and what does that mean about $\lim_{z \rightarrow z_0} (z - z_0)^n g(z)$ ....?
Here is what I think : if $g(z)$ has a pole of order $n$ then the principal part of the Laurent series has ... $n$ terms?
Or has a term to the $-n$ th power?
So that the singularity of $g(z)$ must disappear when multiplied by $(z-z_0)^n$ and hence $\lim_{z \rightarrow z_0} (z - z_0)^n g(z)$ exists and is different from $\pm \infty$.

7. Originally Posted by arbolis
Here is what I think : if $g(z)$ has a pole of order $n$ then the principal part of the Laurent series has ... $n$ terms?
Or has a term to the $-n$ th power?
So that the singularity of $g(z)$ must disappear when multiplied by $(z-z_0)^n$ and hence $\lim_{z \rightarrow z_0} (z - z_0)^n g(z)$ exists and is different from $\pm \infty$.
Correct.