Isolated singularities (Complex Analysis)

**shadow_2145** · October 9th 2008, 10:44 AM

Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity. If the singularity is removable, give the vlaue of the function at the point; if the singularity is a pole, give the order of the pole.

(1) $\frac{e^z-1}{z}$

(2) $\frac{z^4-2z^2+1}{(z-1)^2}$

(3) $\frac{2z+1}{z+2}$

If anyone could show me how to do any of these, I would appreciate it. I don't understand it..Thanks!

**shawsend** · October 9th 2008, 03:10 PM

(1) If $z_0$ is a removable singularity, then $\lim_{z\to z_0} (z-z_0)f(z)=0$ right?

So $\lim_{z\to 0}z\frac{e^z-1}{z}=0$ .

(2) If $f(z)$ has a pole of order $m$ then $f(z)=\frac{\phi(z)}{(z-z_0)^m}$ with $\phi(z_0)\ne 0$ . The numerator of the second one is zero at z=1. So it's not as simple as it looks. Need to factor the numerator:

$\frac{z^4-2z^2+1}{(z-1)^2}=\frac{(z+1)^2(z-1)^2}{(z-1)^2}$ . That just looks like a removable singularity right but we could check that by (1) above:

$\lim_{z\to 1}(z-1)\frac{(z+1)^2(z-1)^2}{(z-1)^2}=0$

The third one has a pole of order 1 at z=-2 by (2).

**mr fantastic** · October 9th 2008, 03:15 PM

Originally Posted by shadow_2145

Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity. If the singularity is removable, give the vlaue of the function at the point; if the singularity is a pole, give the order of the pole.

(1) $\frac{e^z-1}{z}$

Mr F says: Removable singularity at z = 0. (Expand e^z, simplify numerator and note that z is a common factor of numerator and denominator .....) Define f(0) = 1.

(2) $\frac{z^4-2z^2+1}{(z-1)^2}$

Mr F says: Removable singularity at z = 1. (Factorise the numerator .....) Define f(1) = 4.

(3) $\frac{2z+1}{z+2}$

Mr F says: Simple pole at z = -2 because (z + 2) f(z) is differentiable.

If anyone could show me how to do any of these, I would appreciate it. I don't understand it..Thanks!

In each case it's a simple matter of applying the basic definitions (both theortical and practical .....)

**shadow_2145** · November 16th 2008, 08:40 AM

I am reviewing for an exam and I have a problem I don't know how to do.

$\pi cot\pi z$

Any help would be appreciated, thanks!

**mr fantastic** · November 16th 2008, 11:27 AM

Originally Posted by shadow_2145

I am reviewing for an exam and I have a problem I don't know how to do.

$\pi cot\pi z$

Any help would be appreciated, thanks!

You have not said what you don't know how to do here.

$\frac{\pi \, \cos(\pi z)}{\sin (\pi z)}$ obviously has isolated singular points at $z = n$ .

Note that $\lim_{z \rightarrow n} \frac{\pi \, \cos(\pi z) \, (z - n)}{\sin (\pi z)} = 1$ .

Therefore $\frac{\pi \, \cos(\pi z)}{\sin (\pi z)}$ has simple poles at $z = n \, ....$

By the way, new questions should be posted in a new thread.

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