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In Rouche's theorem, If we replace analytic property of functions $f(z)$ and $h(z)$ with meromorphic then this theorem will not be valid anymore.

I want to illustrate this fact by producing some $f(z)$ and $h(z)$ which are meromorphic on some bounded domain D (where D have piecewise smooth boundary $\partial D$). Such that $f(z)$ and $h(z)$ have no poles on $\partial D$ and $|h(z)|<|f(z)|$, $\forall z\in\partial D$ but $f+h$ and $h$ have different number of zeros in D.

mrf
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Sara Tancredi
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    You are right we cannot take away analyliticity in rouches theorem. Because if we do so, we cannot apply The Argument Principle in the proof of the theorem. Rouches theorem is somewhere an consequence of Argument principle. – Mat He Mat Cian Oct 30 '13 at 18:26
  • I'm not sure what you mean it won't be valid anymore. See for example http://math.fullerton.edu/mathews/c2003/rouchetheorem/RoucheTheoremTheorem.3.pdf – Euler....IS_ALIVE Oct 30 '13 at 18:34
  • @ Euler, I mean that it will not imply that zeros of f(z) are equal to zeros of f(z) + h(z). – Sara Tancredi Oct 31 '13 at 01:31

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(I assume you mean $f+h$ and $f$.) Take $D$ the closed unit disc, $h(z) = z^{-1}$ and $f(z) = 2$.

WimC
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