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Problem 19

# Suppose $$\left\\{f_{n}(z)\right\\}$$ is a sequence of analytic functions that converge uniformly to $$f(z)$$ on all compact subsets of a domain $$D$$. Let $$f_{n}\left(z_{n}\right)=0$$ for every $$n$$, where each $$z_{n}$$ belongs to $$D .$$ Show that every limit point of $$\left\\{z_{n}\right\\}$$ that belongs to $$D$$ is a zero of $$f(z)$$.

Expert verified
The limit point $$a$$ of the sequence $$\{z_{n}\}$$ that belongs to the domain $$D$$ is a zero of the function $$f(z)$$ because every subsequence $$\{z_{n_k}\}$$ converging to $$a$$ satisfies $$f_{n_k}(z_{n_k}) = 0$$ due to the uniform convergence of $$\{f_n(z)\}$$ to $$f(z)$$ over the domain $$D$$.
See the step by step solution

## Step 1: Assume a limit point and a converging subsequence

Let's suppose $$a$$ is a limit point of the sequence $$\{z_{n}\}$$, and $$\{z_{n_k}\}$$ is a subsequence that converges to $$a$$ as $$k$$ tends to infinity. By the given conditions, each $$z_{n_k}$$ belongs to the domain $$D$$ and $$f_{n_k}(z_{n_k}) = 0$$ for every $$k$$.

## Step 2: Apply uniform convergence

Since $$\{f_n(z)\}$$ uniformly converges to $$f(z)$$ on all compact subsets of $$D$$, in particular, it converges pointwise to $$f(z)$$ at the point $$a$$. By the property of pointwise convergence, we have $$\lim_{k→∞} f_{n_k}(z_{n_k}) = f(a)$$. Based on Step 1, we have $$f_{n_k}(z_{n_k}) = 0$$ for all $$k$$, hence its limit as $$k$$ tends to infinity also equals 0.

## Step 3: Conclude that limit point is a zero of $$f(z)$$

From Step 2, we get $$f(a) = 0$$. Since $$a$$ is a limit point of $$\{z_n\}$$ that belongs to $$D$$, and given that $$f(a) = 0$$, we can conclude that every limit point of the sequence $$\{z_n\}$$ that belongs to $$D$$ is a zero of $$f(z)$$, which completes the proof.

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