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Could someone advise me on how to approach this problem: Suppose an entire function $f$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. without the use of argument principle ?

Here is my attempt: Suppose $f(z)$ has two zeroes at $z=a.$ Let $f(z) = \sum^{\infty}_{n=0}a_n(z-a)^n, \forall z. \ $

Then $f(z)=(z-a)^2 \left(\dfrac{a_{0}}{(z-a)^2} +\dfrac{a_1}{z-a}+a_2+a_3(z-a)+...\right)$

$\implies a_0=a_1=0.$

$\implies f(z)= (z-a)^2\left(a_2+a_3(z-a)+a_4(z-a)^2+...\right)$

$\implies .... ?$

Thank you.

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Alexy Vincenzo
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1 Answers1

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Let $g:\mathbb R\to\mathbb R$ be the restriction of $f$ to the real line. It suffices to show that $g'$ never vanishes. Suppose to the contrary that $g'(a)=0$. Then $f'(a)=0$, which allows us to write $$f(z)-f(a) = h(z)^k\tag1$$ in a neighborhood of $a$, where $h$ is holomorphic and $k>1$ is the order of zero of $f(z)-f(a)$ at $a$.

Near $a$ we have $h(z) = c(z-a) + o(|z-a|)$ with $c\ne 0$. Hence, $h$ is locally invertible near $a$. The inverse of $h$ transforms the lines $\{te^{\pi ij/k}: t\in\mathbb R\}$, $j=0,\dots, k-1$, to $k$ smooth curves intersecting at $a$. The function $h(z)^k$ is real on each of these curves. In view of (1) this contradicts the assumption that $f(z)$ is real only when $z$ is real.

  • Thanks for the detailed response. May I know how local invertibility of $h$ implies $h(z)^{k}$ is real on $k$ smooth curve? What do you mean by "$k$ smooth curve"? – Alexy Vincenzo Aug 30 '14 at 05:48
  • @AlexyVincenzo I meant "curves". Edited and expanded the answer. –  Aug 30 '14 at 17:41
  • Thanks for your response. I am still quite confused. Why do the lines lie in the domain of $h^{-1} \ ?$ And why are these $k$ curves complex-valued? How do these imply that $h(z)^k$ is real valued on these $k$ curves? – Alexy Vincenzo Aug 31 '14 at 13:36