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Let $f$ be analytic at $z_0 \in \mathbb{C}$ and $f(z_0)=w_0$. Suppose that $f(z)-w_0$ has a zero of finite order in $m \geq 1$ at $z=z_0$.

We need to show that there exist $\epsilon >0$ and $\delta >0$ such that for all $\alpha \in \mathbb{C}$ with $|\alpha -w_0|<\epsilon$, $f(z)-\alpha$ has exactly $m$ simple roots in $B(z_0,\delta)$.

Looking for some Hints. I wish to use to use $$\frac{1}{2\pi i}\int_{\mathcal{C}}\frac{f'(z)}{f(z)}dz = \text{number of zeros of }f \text{ in } Int(\mathcal{C})$$.

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  • $\begingroup$ $\alpha = w_0$?? $\endgroup$ Aug 3, 2018 at 9:26
  • $\begingroup$ you still have $\alpha = w_0$ $\endgroup$ Aug 3, 2018 at 9:28
  • $\begingroup$ just corrected it $\endgroup$
    – Arindam
    Aug 3, 2018 at 9:28
  • $\begingroup$ was my answer below helpful? $\endgroup$ Aug 5, 2018 at 9:26
  • $\begingroup$ Yes, thanks. It was helpful. $\endgroup$
    – Arindam
    Aug 5, 2018 at 16:45

2 Answers 2

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I assume that you have the condition $0<|\alpha -w_0|<\epsilon$; otherwise, as mathworker21 noted, $\alpha=w_0$ will cause trouble. The problem also inherently assumes that $f$ is a nonconstant function.

From the condition that $z=z_0$ is a root of $f(z)-w_0$ of order $m$, we can write $$f(z)=w_0+(z-z_0)^m\,g(z)\,,$$ where $g$ is a holomorphic function and $g(z_0)\neq 0$. Since $g(z_0)\neq0$, there exists a positive real number $\delta_0$ such that, on the neighborhood $B_{2\delta_0}(z_0)$ of $z_0$, $g$ does not vanish. Write $M>0$ for the minimum value of $|g|$ on the boundary $\partial B_{\delta_0}(z_0)$. Write $K$ for the maximum value of $|g'|$ on $\partial B_{\delta_0}(z_0)$. Set $$\delta:=\min\left\{\delta_0,\frac{M}{m\,(K+1)}\right\}>0\,.$$

We first claim that, on $B_{\delta}(z_0)$, $f'(z)$ has at most one root $z=z_0$. (This root has multiplicity $m-1$.) To see this, we note that $$\begin{align} f'(z)&=(z-z_0)^m\,g'(z)+m\,(z-z_0)^{m-1}\,g(z) \\&=(z-z_0)^{m-1}\,\big(m\,(z-z_0)\,g'(z)+g(z)\big)\,. \end{align}$$ If $z\in B_{\delta}(z_0)\setminus\{z_0\}$, then $$\begin{align}\frac{\big|f'(z)\big|}{|z-z_0|^{m-1}}&=\big|m\,(z-z_0)\,g'(z)+g(z)\big|\geq \big|g(z)\big|-m\,\big|z-z_0\big|\,\big|g'(z)\big| \\ &\geq M-m\,\delta\, K\geq M-\left(\frac{M}{K+1}\right)\,K=\frac{M}{K+1}>0\,. \end{align}$$ The claim is now established.

Set $\epsilon:=\delta^m M>0$. For $\alpha \in B_{\delta}(z_0)$, we see that $$f(z)-\alpha=(z-z_0)^m\,g(z)-(\alpha-w_0)\,.$$ When $\alpha\in B_{\epsilon}(w_0)$ and $z\in\partial B_{\delta}(z_0)$, we have $$\big|(z-z_0)^m\,g(z)\big|\geq \delta^m M=\epsilon> |\alpha-w_0|\,.$$ Using Rouché's Theorem, the number of roots of $f(z)-\alpha$ in $B_{\delta}(z_0)$ is the same as the number of roots of $(z-z_0)^mg(z)$ in $B_{\delta}(z_0)$, which is $m$.

We now prove that, if $0<|\alpha-w_0|<\epsilon$, then $f(z)-\alpha$ has no repeated roots inside $B_{\delta}(z_0)$. First, by the choice of $\delta$, the only roots of $f'(z)$ inside $B_{\delta}(z_0)$ is $z_0$. Second, if $z=\zeta\in B_{\delta}(z_0)$ is a double root of $f(z)-\alpha$, then both $f(\zeta)-\alpha$ and $\left(\dfrac{\text{d}}{\text{d}z}\,\big(f(z)-\alpha\big)\right)\Bigg|_{z=\zeta}=f'(\zeta)$ must be both $0$. However, $f'(\zeta)=0$ implies $\zeta=z_0$, but then $f(z_0)-\alpha=w_0-\alpha\neq 0$. This is a contradiction, whence such $\zeta \in B_{\delta}(z_0)$ does not exist.

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Well, you can use Rouché's theorem, which follows from the equation you wrote. $f(z)-\alpha = (f(z)-w_0)-(w_0-\alpha)$. You can take $\delta > 0$ so that $z_0$ is the only root of $f$ in $B(z_0,2\delta)$. Then take $\epsilon = \frac{1}{2}\sup_{z \in \partial B(z_0,\delta)} |f(z)-w_0|$, so that $|w_0-\alpha| \le \epsilon < |f(z)-w_0|$ for every $z \in \partial B(z_0,\delta)$.

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