# Removable singularity

• March 9th 2011, 04:23 AM
skyking
Removable singularity
Let $f$ be analytic in $\{ z:0<|z-\alpha | and has a singularity at $\alpha$. Supppose there exists a neighbourhood $\{ z:0<|z-\alpha |<\epsilon \}$ where $Re(f(z))>0$. Show that $\alpha$ is a removable singularity.

I've tried looking at $e^{f(z)}$ in order to show that $f$ is bounded at a neighbourhood of $\alpha$, but all I get is $|e^{f(z)}|=e^{Re(f(z))}>1$ so that dosent help, I also looked at $e^{-f(z)}$ but got nowhere. Need some direction here please.
• March 9th 2011, 06:55 AM
Tinyboss
You know it's not essential, because an analytic function attains almost every value in any neighborhood of an essential singularity. So you have to rule out a pole. So suppose there's a pole at alpha, and consider $[f(z)]^{-1}$. It has a removable singularity: $h(z)=[f(z)]^{-1}$ for $z\ne\alpha$ and $h(\alpha)=0$ is analytic on the original domain.

What does its image look like?
• March 9th 2011, 07:33 AM
FernandoRevilla
Quote:

Originally Posted by Tinyboss
You know it's not essential, because an analytic function attains almost every value in any neighborhood of an essential singularity.

Good, an elegant application of the Casaroti-Weierstrass theorem.
• March 9th 2011, 07:34 AM
skyking
Quote:

Originally Posted by Tinyboss
You know it's not essential, because an analytic function attains almost every value in any neighborhood of an essential singularity. So you have to rule out a pole. So suppose there's a pole at alpha, and consider $[f(z)]^{-1}$. It has a removable singularity: $h(z)=[f(z)]^{-1}$ for $z\ne\alpha$ and $h(\alpha)=0$ is analytic on the original domain.

What does its image look like?

I'm not sure I understand why $h$ is analytic. We know $g(z)=|z|$ is not analytic at $z=0$ ?

It's image is the positive half of the real line, including zero. How does that contradict $\alpha$ being a pole?
• March 9th 2011, 08:05 PM
Tinyboss
I was a little sloppy--h won't be analytic on all the original domain, but only where f is nonzero. But since we're assuming f has a pole at alpha, then f is nonzero on some neighborhood. If the real part of z is positive, then so is the real part of 1/z. But zero is also in the image of h, violating the open mapping property of analytic maps.
• March 10th 2011, 07:21 AM
skyking
I actually mis-spoke earlier. Checked the book again and $|z|$ is not analytic in the entire complex plane.

So now I'm completely confused (Headbang)
• March 10th 2011, 08:42 AM
chisigma
Quote:

Originally Posted by skyking
Let $f$ be analytic in $\{ z:0<|z-\alpha | and has a singularity at $\alpha$. Supppose there exists a neighbourhood $\{ z:0<|z-\alpha |<\epsilon \}$ where $Re(f(z))>0$. Show that $\alpha$ is a removable singularity.

I've tried looking at $e^{f(z)}$ in order to show that $f$ is bounded at a neighbourhood of $\alpha$, but all I get is $|e^{f(z)}|=e^{Re(f(z))}>1$ so that dosent help, I also looked at $e^{-f(z)}$ but got nowhere. Need some direction here please.

A 'counterexample': let's suppose $f(z)= \ln \frac{1}{z}$... setting $z= r\ e^{i\ \varphi}$ is $f(z)= - \ln r - i\ \varphi$ so that in $0<|z|<1$ $f(z)$ is analytic and is also $\text {Re} \{f(z)\} >0$ ... but , in $z=0$ , $f(z)$ certainly doesn't have a 'removable singularity'...

Kind regards

$\chi$ $\sigma$
• March 11th 2011, 06:38 AM
skyking
Quote:

Originally Posted by chisigma
A 'counterexample': let's suppose $f(z)= \ln \frac{1}{z}$... setting $z= r\ e^{i\ \varphi}$ is $f(z)= - \ln r - i\ \varphi$ so that in $0<|z|<1$ $f(z)$ is analytic
Kind regards

$\chi$ $\sigma$

But, $Logz$ is not even defined for negative values, and in your 'counterexample' $\frac {1}{z}$ clearly takes real negative values in $0<|z|<1$ so I think this example is not valid.

Still need help showing the above, still stuck
• March 11th 2011, 09:37 AM
chisigma
Quote:

Originally Posted by skyking
But, $Logz$ is not even defined for negative values, and in your 'counterexample' $\frac {1}{z}$ clearly takes real negative values in $0<|z|<1$ so I think this example is not valid.

Still need help showing the above, still stuck

Just for curiosity: where did You read that?... perhasps in the Boys Scout Manual?... $z$ is a complex variable that can be written, according to De Moivre's formula, as $z= r\ e^{i\ \varphi}$ so that the [complex] logarithm $\ln z = \ln r + i\ \varphi$ is defined in the entire $\mathbb{C}$ with the only exception of $z=0$...

Kind regards

$\chi$ $\sigma$
• March 11th 2011, 10:18 AM
Opalg
Quote:

Originally Posted by chisigma
$z$ is a complex variable that can be written, according to De Moivre's formula, as $z= r\ e^{i\ \varphi}$ so that the [complex] logarithm $\ln z = \ln r + i\ \varphi$ is defined in the entire $\mathbb{C}$ with the only exception of $z=0$...

You have to be very careful here, because the complex logarithm is not a single-valued function in general. It depends on the argument $\varphi$, which is only defined up to a multiple of $2\pi$. In particular, the function $\ln\frac1z$ cannot be defined in a way that makes it continuous in a (punctured) neighbourhood of the origin. So this function cannot be analytic in any such neighbourhood.
• March 11th 2011, 11:02 AM
chisigma
Quote:

Originally Posted by Opalg
You have to be very careful here, because the complex logarithm is not a single-valued function in general. It depends on the argument $varphi$, which is only defined up to a multiple of $2pi$. In particular, the function $lnfrac1z$ cannot be defined in a way that makes it continuous in a (punctured) neighbourhood of the origin. So this function cannot be analytic in any such neighbourhood.

The question has been investigated in...

http://www.mathhelpforum.com/math-he...st-167358.html

... and here it has been demonstrated, using the Cauchy-Riemann relations, that $\ln z$ is analytic everywhere in $\mathbb{C}$ with the only exception of $z=0$. Regarding the question proposed by skyking I suggest, in order to avoid problems, to modify the first line as follows...

Let $f(z)$ be analytic in $\{z : 0<|z-\alpha|< r \}$, has a singularity at $\alpha$ and $\alpha$ is not a branch point of $f(z)$ ...

Logarithmic Branch Point -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$
• March 11th 2011, 11:20 AM
chisigma
Quote:

Originally Posted by Opalg
You have to be very careful here, because the complex logarithm is not a single-valued function in general. It depends on the argument $\varphi$, which is only defined up to a multiple of $2\pi$. In particular, the function $\ln\frac1z$ cannot be defined in a way that makes it continuous in a (punctured) neighbourhood of the origin. So this function cannot be analytic in any such neighbourhood.

Of course is $\ln \frac{1}{z} = - \ln z$ so that what holds for $\ln z$ holds for $-\ln z$ too...

Kind regards

$\chi$ $\sigma$
• March 12th 2011, 06:54 PM
skyking
Quote:

Originally Posted by chisigma
Just for curiosity: where did You read that?... perhasps in the Boys Scout Manual?...

Kind regards

$\chi$ $\sigma$

Very classy...:rolleyes: