I have this question which I struggled to apply Rouche's Theorem:

Let $f$ and $g$ be analytic inside and on a smooth regular closed curve $\gamma$. Suppose that $f(z) \neq 0$ for all $z$ on $\gamma$. Prove that there is $\varepsilon > 0$ such that $\mathbb{Z}(f) = \mathbb{Z}(f +\varepsilon g)$ inside $\gamma$, where $\mathbb{Z}$ is the number of zeros.

Any help or hints is appreciated.

  • $\begingroup$ @KennyWong yes my mistake so sorry it should be not equal. $\endgroup$
    – Noob4398
    Commented May 19, 2017 at 8:38
  • 2
    $\begingroup$ I suppose that the hypothesis is $f(z)\not =0$ on $A=\{\gamma(t), t\in [0,1]\}$. Then as $A$ is compact, there exist $m>0$ such that $|f(z)|\geq m$ for $z$ in $A$, and $M$ such that $|g(z)|\leq M$ for $z\in A$. Now for $\epsilon$ close to $0$, you have $\varepsilon M<m$, so you can apply Rouché. $\endgroup$
    – Kelenner
    Commented May 19, 2017 at 8:40
  • $\begingroup$ @Kelenner Is the compactness due to the Heine-Borel theorem, and the bounds on f and g by the Extreme Value Theorem? $\endgroup$
    – Noob4398
    Commented May 19, 2017 at 10:44
  • $\begingroup$ $A$ is the image of $[0,1]$ compact by $\gamma$ continuous, hence compact. On a compact a continuous function (here $|f(z)|$ or $|g(z)|$) is bounded and the bound are realized. $\endgroup$
    – Kelenner
    Commented May 19, 2017 at 11:57


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