Google Answers Logo
View Question
Q: conformal mapping ( Answered,   0 Comments )
Subject: conformal mapping
Category: Science > Math
Asked by: fmunshi-ga
List Price: $12.00
Posted: 09 Dec 2002 18:01 PST
Expires: 08 Jan 2003 18:01 PST
Question ID: 122126
Prove that if f has a simple pole at zo then there exists a punctured
neighborhood of zo on which f is one-to-one
Subject: Re: conformal mapping
Answered By: dannidin-ga on 10 Dec 2002 01:02 PST
Hi fmunshi,

I assume that you mean f is analytic in a punctured neighborhood of z0
and has a simple pole at z0.

Therefore we may write f(z) = 1/(z-z0) + g(z), where g(z) is analytic
and bounded in a punctured neighborhood of z0, and therefore analytic
in a (non-punctured) neighborhood of z0, by the Riemann bounded
singularity theorem
(or by Taylor series).

If z1 and z2 are such that f(z1)=f(z2), then

g(z2) - g(z1) = 1/(z1-z0) - 1/(z2-z0) = (z2 - z1) / ((z1-z0)(z2-z0)),

(g(z2)-g(z1))/(z2-z1) = 1/((z1-z0)(z2-z0))

Taking absolute values we get

|g(z2)-g(z1)|          1
|-----------| = ---------------
|  z2-z1    |   |z1-z0|*|z2-z0|

But now, this cannot happen if z1 and z2 are sufficiently close to z0,
because if |z1-z0|, |z2-z0| is < epsilon, then the right hand side of
the above equation is at least 1/epsilon^2. However, the left hand
side must remain bounded because for z1,z2 in some punctured
environment U of z0,

|-----------|  < or =    sup   |g'(z)|
|   z2-z1   |           z in U

(to prove this inequality, write g(z2)-g(z1) as integral from z1 to z2
of g'(z), then take absolute values and use the triangle inequality
for integrals)

In other words, now to formulate a precise proof that f is one-to-one
in some punctured neighborhood of z0:

Let M be a bound of |g'(z)| in some punctured neighborhood of z0. Let
epsilon be small enough so that 1/epsilon^2 > M (and small enough so
that f and g are defined in the disc D(z0,epsilon) of radius epsilon
around z0). Then for any z1 and z2 in D(z0,epsilon)\{z0}, we cannot
have f(z1)=f(z2) due to the above inequality.

Hope this helps. If any of this is unclear or too brief please ask and
I will gladly clarify.

There are no comments at this time.

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  

Google Home - Answers FAQ - Terms of Service - Privacy Policy