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)),
or
(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,
|g(z2)-g(z1)|
|-----------| < 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.
Regards,
dannidin |
