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Re: Bracket a complex root
Posted:
Jun 18, 1996 7:32 PM


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 xvalues for > which the yvalues 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



