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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 \(f(z)\).

Short Answer

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 \(D\).
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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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