P.106, Ch3, Q16 in Complex Analysis, Stein

Suppose $f$ and $g$ are holomorphic in a region containing the disc $|z|\leq 1$. Suppose that $f$ has a simple zero at $z=0$ and vanishes nowhere else in $|z|\leq 1$. 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 $|z|\leq 1$, and

(b) if $z_{\epsilon}$ is this zero, the mapping $\epsilon\mapsto z_{\epsilon}$ is continuous.

Would like to know how to do (b).

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2 Responses to P.106, Ch3, Q16 in Complex Analysis, Stein

1. KK Kwong says:

(I can’t post the comment properly so I tried to break it into two. )

(a) $|f_\epsilon -f|=\epsilon |g|<|f|$ on the circle $\{|z|=1\}$ if $\epsilon$ is small enough, so by Rouche’s theorem $f_\epsilon$ has a unique zero in the unit disk.

2. KK Kwong says:

(b) I only prove that it is continuous at $\epsilon=0$, the general case is the same (by a “translation” argument) but only notationally more difficult.
Let’s denote $h(\mu)=z_{\mu}$, clearly $h(0)=0$. Now, suppose there exists $\epsilon>0$ such that for all $n$, there is a $|\mu_n|<1/n$ with $|h(\mu_n)|=:|z_n|\geq \epsilon$. We can assume that (by taking a subsequence) $z_n\rightarrow z_0$ with $1\geq |z_0|\geq \epsilon$. Then as $f(z_n)=-\mu_n g(z_n)$, by letting $n\rightarrow \infty$, we have $f(z_0)=0$, a contradiction.