# Simple Pole

1. Apr 8, 2007

### Diophantus

How exactly would one go about proving that

$$\frac{1}{1-2^{1-z}}$$ has a simple pole at $$z=1$$?

I've tried writing $$2^{1-z}$$ in terms of e to get a Taylor series for the denominator but can't quite figure out where to go from there.

2. Apr 8, 2007

### Gib Z

In complex analysis, a pole of a holomorphic function is a certain type of simple singularity that behaves like the singularity 1/z^n at z = 0. A pole of order 1 is called a simple pole as you know. Essentially you need to prove
z behaves like 1-2^(1-z) near zero. Perhaps if you showed that $$\lim_{z\to 0} \frac{z}{1-2^{1-z}} = 1$$...

3. Apr 8, 2007

### Gib Z

Scratch that, im not sure that helps..All you should need to show is that it approaches the same limit, so the first one, z, as z goes to 0, its just zero. the second one is also 0. That should do it.

If your not happy, show that the function Laurent series near z=1 below degree −n vanishes and the term in degree −n is not zero.

Last edited: Apr 8, 2007
4. Apr 8, 2007

### Diophantus

OK so I've opted for the Laurant Series route. Where am I going wrong here:

$$\frac{1}{1-2^{1-z}} = \displaystyle\sum_{n=0}^{\infty}(2^{1-z})^n = \displaystyle\sum_{n=0}^{\infty}(e^{(1-z)log2})^n = \displaystyle\sum_{n=0}^{\infty}(\displaystyle\sum_{m=0}^{\infty}\frac{(1-z)^m(log2)^m}{m!})^n$$

This doesn't look like it has any singilarities.

Last edited: Apr 9, 2007
5. Apr 10, 2007

### Zurtex

I'm not sure using the geometric series is really valid for about z = 1, I can get an appropriate series on mathematica which shows it is a simple pole, but I have no idea how to generate it myself.

6. Apr 10, 2007

### marcusl

If it has a simple pole, then the residue should be bounded and given by

$$R=\lim_{z\rightarrow 1} (z-1)f(z)=\lim_{z\rightarrow 1} \frac{z-1}{1-2^{1-z}}.$$

Substitute $$u=z-1$$ to get

$$R=\lim_{u\rightarrow 0} \frac{u}{1-2^{-u}}$$

Evaluating using l'Hospital's rule gives R=2.
EDIT: Accordingly, 2/(z-1) is a term in the Laurent series.
EDIT2: define u=z-1 instead of 1-z for clarity

Last edited: Apr 10, 2007
7. Apr 10, 2007

### Diophantus

Hmmm, it's actually 1/log2 but thanks anyway. I'm really intrigued to know whether there is a reasonable way of calculating the Laurant series though. Surely there must be a bit of trickery that will work.

8. Apr 10, 2007

### marcusl

Oops! I took the derivative wrong! Very sorry

9. Apr 10, 2007

### mathwonk

you want to show [z-1] over that function is bounded.

but you just want to show 2^(1-z) equals 1 simply when z=1, or that 2^z equals 1 simply when z=0, but thats sort of clear, since the derivative is not zero anywhere, 2^z has no multiple values at all.

10. Apr 11, 2007

### marcusl

I'll repeat my earlier derivation, without the silly error.

$$f(z)=\frac{1}{1-2^{1-z}}=\frac{2^z}{2^z-2}$$

has a pole if the residue

$$R=\lim_{z\rightarrow 1} (z-1)f(z)=\lim_{z\rightarrow 1} \frac{2^z (z-1)}{2^z-2}$$.

is bounded. To evaluate via l'Hospital's rule, write

$$2^z=(e^{\ln2})^z=e^{z\ln2}$$

so the derivative is

$$\frac{d(2^z)}{dz}=2^z \ln(2).$$

Then

$$R=\lim_{z\rightarrow 1} \frac{2^z[1+(z-1)\ln(2)]}{2^z \ln2} = \frac{1}{\ln(2)} .$$

Hope I redeemed myself!

EDIT: It's even simpler to define u=z-1, then

$$R=\lim_{u\rightarrow 0} \frac{u}{1-2^{-u}}$$

and applying l'Hospital's rule gives

$$R=\lim_{u\rightarrow 0} \frac{1}{2^u \ln(2)}=\frac{1}{\ln(2)}.$$

Last edited: Apr 11, 2007
11. Apr 11, 2007

### mathwonk

didnt i make this trivial? or did i screw up?

12. Apr 12, 2007

### marcusl

Can't answer 'cause I don't know what you mean by "2^(1-z) equals 1 simply when z=1"?

Last edited: Apr 12, 2007
13. Apr 13, 2007

### Gib Z

>.<" Perhaps the "simply" is troubling you? Thats not a mathematical term he's using, hes just saying that it is simple.

For 2^n to equal 1, the only value n can be is 0. In this case, n is 1-z.
1-z=0. z=1.

14. Apr 13, 2007

### Diophantus

What about $$z=1 + \frac{2\pi in}{log(2)}$$ for any integer n?

Last edited: Apr 13, 2007
15. Apr 13, 2007

### Gib Z

No need for that double post :P And we are dealing with the primary branch only, otherwise nothing we talk about here are functions anymore.

16. Apr 13, 2007

### Diophantus

I'm sorry you've lost me now. Is it or is it not true that the function 2^(1-z) has poles at $$z=1 + \frac{2\pi in}{log(2)}$$?

What do you mean by primary branch?

17. Apr 13, 2007

### Diophantus

Oh I see now, the primary branch of the complex log function. I forgot about that.

Still, do we have poles here or not?

18. Apr 13, 2007

### Gib Z

Yes! Think about the definiton of a simple pole, and what mathwonk said.

19. Apr 13, 2007

### Diophantus

I don't understand why you said

when it is clearly wrong.

And what's all this talk about it not being a function? I have only ever alluded to 2^n being 1 whose imaginary part, if I am not mistaken, lies between + and - pi.

20. Apr 14, 2007

### Gib Z

"For 2^n to equal 1, the only value n can be is 0. In this case, n is 1-z.
1-z=0. z=1. "

I don't understand why that is clearly wrong...