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

Asked by Ho Pak.

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.

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