Contour Integrals

The Observer · **Joined:** Thu, 31 May 2012 03:22:29 UTC **Posts:** 6

Hello, I read somewhere in a book that the number of zeros N and the number of poles P of a function within contour C can be found by
$N-P=\frac{1}{2\pi i} \oint_{C} \frac{f'\left( {z} \right) dz}{f\left( {z} \right)}$
This is easy to see because you can find resides of functions in the form f(z)/g(z) by f(z)/g'(z), so f'(z)/f'(z)=1, but when I try to evaluate integrals like this, the results do not match those of the residue theorem. Does anyone here have examples of this with steps?

Thanks!! :confused:

outermeasure · **Posted:** Fri, 17 Aug 2012 03:44:26 UTC

The Observer wrote:

Hello, I read somewhere in a book that the number of zeros N and the number of poles P of a function within contour C can be found by
$N-P=\frac{1}{2\pi i} \oint_{C} \frac{f'\left( {z} \right) dz}{f\left( {z} \right)}$
This is easy to see because you can find resides of functions in the form f(z)/g(z) by f(z)/g'(z), so f'(z)/f'(z)=1, but when I try to evaluate integrals like this, the results do not match those of the residue theorem. Does anyone here have examples of this with steps?

Thanks!! :confused:

What do you mean by "the results do not match those of the residue theorem"?

The Observer · **Joined:** Thu, 31 May 2012 03:22:29 UTC **Posts:** 6

If I evaluate them with the residue theorem, I get the expected result, but not when I just evaluate the contour integral. It's probably something very simple that I'm missing, that's why I wanted to see a worked example.

Shadow · **Posted:** Fri, 17 Aug 2012 06:18:32 UTC

The Observer wrote:

If I evaluate them with the residue theorem, I get the expected result, but not when I just evaluate the contour integral. It's probably something very simple that I'm missing, that's why I wanted to see a worked example.

Huh? That's the argument principle.

outermeasure · **Posted:** Fri, 17 Aug 2012 07:36:40 UTC

The Observer wrote:

If I evaluate them with the residue theorem, I get the expected result, but not when I just evaluate the contour integral. It's probably something very simple that I'm missing, that's why I wanted to see a worked example.

This is probably the simplest example: Let $f\equiv 1$ . Then $f'\equiv 0$ , N=P=0

, $\displaystyle\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}\,\mathrm{d}z=0$ whatever contour

is chosen.

The Observer · **Joined:** Thu, 31 May 2012 03:22:29 UTC **Posts:** 6

I know this is the argument principle. Does anyone have any examples that do have zeros? Preferably using a special function. I've been digging through my complex analysis books and I'm finding no useful examples that don't use the residue theorem.

Shadow · **Posted:** Sat, 18 Aug 2012 00:31:04 UTC

The Observer wrote:

I know this is the argument principle. Does anyone have any examples that do have zeros? Preferably using a special function. I've been digging through my complex analysis books and I'm finding no useful examples that don't use the residue theorem.

If you want something more specific you should clarify exactly what it is you're having trouble with. I'm still not sure I understand your original question. Can you give an example where you've worked out the residue theorem and some other way and they do not match?

The Observer · **Joined:** Thu, 31 May 2012 03:22:29 UTC **Posts:** 6

I am talking about a function f(z) with N zeros and P poles. f'(z)/f(z) does not simplify and must be integrated in that form without using the residue theorem to evaluate the integral. I need to know the easiest way to deal with that. I often see it written as
$N-P=\frac{1}{2\pi}\Delta_{C}arg[f(z)]$
where C is the contour. Can someone explain this and how to evaluate it?

Thanks!!

Shadow · **Posted:** Sat, 18 Aug 2012 22:09:28 UTC

The Observer wrote:

I am talking about a function f(z) with N zeros and P poles. f'(z)/f(z) does not simplify and must be integrated in that form without using the residue theorem to evaluate the integral. I need to know the easiest way to deal with that. I often see it written as
$N-P=\frac{1}{2\pi}\Delta_{C}arg[f(z)]$
where C is the contour. Can someone explain this and how to evaluate it?

Thanks!!

The residue theorem is exactly how you get that, you just need to write out the Taylor or Laurent series at a point where there's a pole or zero and you'll see it come out to exactly the N-P (CM).

outermeasure · **Posted:** Sun, 19 Aug 2012 03:16:31 UTC

Shadow wrote:

The Observer wrote:

I am talking about a function f(z) with N zeros and P poles. f'(z)/f(z) does not simplify and must be integrated in that form without using the residue theorem to evaluate the integral. I need to know the easiest way to deal with that. I often see it written as
$N-P=\frac{1}{2\pi}\Delta_{C}arg[f(z)]$
where C is the contour. Can someone explain this and how to evaluate it?

Thanks!!

The residue theorem is exactly how you get that, you just need to write out the Taylor or Laurent series at a point where there's a pole or zero and you'll see it come out to exactly the N-P (CM).

... or you write $f(z)=g(z)\dfrac{\prod (z-z_j)}{\prod (z-p_k)}$ where the z_j

are the zeros and p_k

are the poles (with multiplicity), and

is nonvanishing holomorphic on some simply-connected domain containing the track of C.

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