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 ```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```
 ```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```