Discussion
Let \(\mathbb{D} = \{z \in \mathbb{C} : |z| \lt 1\}\). Then there exists a non-constant analytic function \(f\) on \(\mathbb{D}\) such that for all \(n = 2, 3, 4, \ldots\)
Answer: C
Such a function exists for the zeros \(f\left(1 - \frac{1}{n}\right) = 0\).
By the identity theorem, the zeros of a non-constant analytic function on \(\mathbb{D}\) cannot accumulate at a point inside \(\mathbb{D}\). The points \(\frac{i}{n}\) and \(\frac{1}{n}\) accumulate at 0, and \(\frac{1}{2} - \frac{1}{n}\) accumulate at \(\frac{1}{2}\), both inside the disc, so those options force \(f \equiv 0\).
The points \(1 - \frac{1}{n}\) accumulate only at 1, on the boundary, which is allowed. An example is
\[ f(z) = \sin\left(\frac{\pi}{1 - z}\right) \]
which is analytic on \(\mathbb{D}\) and vanishes where \(\frac{1}{1-z} = n\), that is at \(z = 1 - \frac{1}{n}\).
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