Suppose \(f\) is analytic on \(\DD^\circ\). Determine with proof which of the following are possible:
\(f\qty{1\over n} = (-1)^n\) for each \(n>1\).
\(f\qty{1\over n} = e^{-n}\) for each even integer \(n>1\) while \(f\qty{1\over n} = 0\) for each odd integer \(n>1\).
\(f\qty{1\over n^2} = {1\over n}\) for each integer \(n>1\).
\(f\qty{1\over n} = {n-2 \over n-1}\) for each integer \(n>1\).
Solution
The points \(1/n\) and \(1/n^2\) accumulate at \(0\in\mathbb D\), so analyticity at \(0\) strongly constrains all four cases.
(a) Impossible. Continuity at \(0\) would force the sequence \(f(1/n)\) to converge to \(f(0)\), but \((-1)^n\) does not converge.
(b) Impossible. For every odd \(n>1\), \(f(1/n)=0\). These zeros accumulate at \(0\in\mathbb D\), so the identity theorem gives \(f\equiv0\). This contradicts \(f(1/n)=e^{-n}\ne0\) for even \(n\).
(c) Impossible. Since \(f(1/n^2)=1/n\to0\), continuity gives \(f(0)=0\). If \(f\) is not identically zero, let \(m\ge1\) be the order of its zero at \(0\). Then \[ f(z)=z^m g(z), \qquad g(0)\ne0. \] At \(z=1/n^2\) this gives \[ \frac1n=\frac{1}{n^{2m}}g(1/n^2), \] so \[ g(1/n^2)=n^{2m-1}, \] which diverges, contradicting \(g(1/n^2)\to g(0)\). The identically-zero case is also impossible.
(d) Possible. Define \[ f(z)=\frac{1-2z}{1-z}. \] Its only pole is at \(z=1\), so it is holomorphic on the open unit disk. For \(n>1\), \[ f(1/n) =\frac{1-2/n}{1-1/n} =\frac{n-2}{n-1}. \]
Hence precisely (d) can occur.