Re: e^z = z
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 02 Apr 2009 14:08:06 -0500
Timothy Murphy <gayleard@xxxxxxxxxx> writes:
Is there a simple way of showing that there exists
a complex number z such that e^z = z?
Eg is there some Newton-like iterative approach to a solution?
Or can one use Rouche's theorem in some way?
The result seems to me to follow from the fact that
if there is no solution then f(z) = e^z - z
is an entire function of order 1 and so f(z) = e^{az + b}
which is easily seen to be impossible.
But I am looking for a solution at a lower level.
Consider the image under f of the rectangle with corners 0, N,
N + 2 pi i, 2 pi i, where N is large.
The image of the segment 0 to N goes along the positive real
axis from 1 to R = exp(N)-N >> N. The image of the segment N to N + 2 pi i
is within distance N+2 pi of a counterclockwise circular arc, looping
around the origin and coming back to R - 2 pi i. The image of the segment
N + 2 pi i to 2 pi i is a straight line from R - 2 pi i to 1 - 2 pi i.
The image of the segment from 2 pi i to 0 goes from 1 - 2 pi i to 1 in
the lower half plane (since sin(t) - t < 0 for t > 0). So this contour has
winding number 1 around the origin, and therefore there is one zero of f(z)
inside the rectangle.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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