I've been working on a generalization of Rouche's theorem proposed by a classmate, he postulated that

"If $h$ and $f$ are holomorphic functions on $H=\{z\in\mathbb{C}:Re(z)>0\}$ such that $|h(z)|<|f(z)|$ for all $z=bi$ with $b\in\mathbb{R}$, then $f$ and $f+h$ has the same quantity of ceros on $H$ counting multiplicity."

To prove this the principal idea was to use the lineal fractional transformation $t(z)=\frac{z+1}{1-z}$ which send the unit circle to $H$, so we have that


on $|z|=1$, and we can conclude using the regular Rouche's theorem.

THE PROBLEM is that there have to be something wrong in this proof, because this result doesn't really holds. Indeed we can take $f(z)=e^z$, $h(z)=\frac{1}{2e^z}$ and observe that clearly $|h(z)|<|f(z)|$ on the imaginary axis ($\frac{1}{2}<1$) but $(h+f)(z)=e^z+\frac{1}{2e^z}$ has zeros while $f$ doesn't has any.

What is bad? Is this conjecture reparable?

Anything would be really helpfull!


1 Answer 1


The basic problem is that $h \circ t$ and $f \circ t$ are not analytic at $z=1$. You would need to make some assumptions about limits of $f$ and $h$ at $\infty$ that would let you restrict attention to a bounded region where the ordinary Rouché's theorem applies.

  • $\begingroup$ Thank you very much! So you say that if, besides the hypothesis in the conjecture, you add that $lim|h(z)|<lim|f(z)|$ we could conclude?? $\endgroup$
    – Esteban G.
    Nov 15, 2016 at 2:45

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