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Topic: Bracket a complex root
Replies: 11   Last Post: Jun 24, 1996 2:12 AM

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hawthorn@waikato.ac.nz

Posts: 14
Registered: 12/12/04
Re: Bracket a complex root
Posted: Jun 18, 1996 7:32 PM
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In article <4q2uts$8n1@btmpjg.god.bel.alcatel.be>, pver@nemdev26 (Peter Verthez) writes:
> Does anyone know how to 'bracket' a complex root ?
>
> Let me explain a bit: when you try to numerically find a real root
> of a continuous function, you would begin by finding two x-values for
> which the y-values have opposite signs. Then you know that there is
> a root in between them.
>
> Now take the complex case, suppose you have an analytic function
> w = f(z) in the complex plane (analytic in some region). Is there a way
> to determine a region (e.g. a square in the complex plane) in which you will
> certainly find a complex root ?


Yep there is - it is called a contour integral,

Let f be a function analytic on the complex plane. Let p be a closed path
on the complex plane. Assume that p does not touch a zero of f.

Form the integral
/
1 | f'(z)
------ | ------- dz
2 pi i | f (z)
/
path p

then this is equal to the sum of the winding numbers of the
path p around the roots (with multiplicities) of the function f.

In particular, you can take p to be the boundary of a square.
The above function will tell you the number of roots (with
multiplicities) which lie inside the square. Neat eh?

Ian H







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