Suppose that $f$ and $g$ are holomorphic in a domain containing the unit disc $D=\{z| |z| \le 1 \}$. Suppose that $f$ has a simple zero at $z=0$ and vanishes nowhere in the unit disc.
Let $f_{\epsilon}(z)=f(z)+ \epsilon g(z)$. Show that if $\epsilon$ is sufficiently small then
a) $f_{\epsilon}(z)$ has a unique zero in the unit disc.
b) Show that the map $\epsilon \to z_{\epsilon}$ is continuous
My try:
For (a): Since $f$ has a simple zero at $z=0$ and vanishes nowhere in the unit disc, on $|z|=1$ ,$f$ attains the minimum value (say $\delta \gt 0$). Since $g$ is holomorphic on $D$, $g$ attains its maximum (say $M$). Then we have $$|f_{\epsilon}(z)-f(z)|=| \epsilon g(z)| \lt \epsilon M$$
For $\epsilon \lt \frac{\delta}{M}$, on $|z|=1$, we have $$|f_{\epsilon}(z)-f(z)| \lt \delta \le |f(z)|$$ . The conclusion follows by Rouche's theorem.
For (b)
I have no idea on how to proceed.
Thanks for the help!!
Best Answer
If a holomorphic function $h$ has a zero of multiplicity $k$ at $z_0$, then we have $h(z) = (z-z_0)^k\cdot \tilde{h}(z)$ with $\tilde{h}$ holomorphic in a neighbourhood of $z_0$ and $\tilde{h}(z_0) \neq 0$. Thus we have
$$\frac{h'(z)}{h(z)} = \frac{k(z-z_0)^{k-1}\tilde{h}(z) + (z-z_0)^k\tilde{h}'(z)}{(z-z_0)^k\tilde{h}(z)} = \frac{k}{z-z_0} + \frac{\tilde{h}'(z)}{\tilde{h}(z)},$$
and the second summand is holomorphic in a neighbourhood of $z_0$. The residue of $\frac{h'}{h}$ in $z_0$ is therefore $k$. Then
$$z\frac{h'(z)}{h(z)} = \frac{k\cdot z}{z-z_0} + z\frac{\tilde{h}'(z)}{\tilde{h}(z)},$$
so
$$\operatorname{Res} \biggl(z\frac{h'}{h}; z_0\biggr) = \operatorname{Res} \biggl(\frac{k\cdot z}{z-z_0}; z_0\biggr) = k\cdot z_0.$$
Here, $f_\epsilon$ has only one zero in the unit disk (counting multiplicities), and therefore
$$z_\epsilon = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} z\cdot\frac{f_\epsilon'(z)}{f_\epsilon(z)}\,dz = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} z\cdot\frac{f'(z) + \epsilon g'(z)}{f(z) + \epsilon g(z)}\,dz,$$
which shows that the function $\epsilon \mapsto z_\epsilon$ is continuous [even holomorphic], since the integrand depends continuously [holomorphically] on $\epsilon$.