Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 4E (Complex Analysis)

Set $h=f-g$, an entire function. By the Bolzano-Weierstrass theorem, a subsequence of the distinct zeros $z_i$ converges to a point $a$ in the closed unit disc. An analytic function has a convergent Taylor series in a neighbourhood of each point. If the first nonzero coefficient of $h$ at $a$ had index $m$, thenwith $k$ analytic and nonzero in a sufficiently small disc. The zero $a$ would then be isolated, contradicting the accumulating distinct zeros. All Taylor coefficients at $a$ vanish, so $h$ vanishes on a disc.

To justify continuation across the whole plane, let $Z$ be the set of points at which $h$ vanishes on a neighbourhood. This set is nonempty and open. It is also closed: if points of $Z$ tend to $b$, continuity of each derivative gives $h^{(m)}(b)=0$ for every $m$, and the Taylor series at $b$ gives a zero neighbourhood. Since $\mathbb{C}$ is a connected space, $Z=\mathbb{C}$. This proves the identity theorem here and gives $\boxed{f\equiv g}$.

Without boundedness, the distinct zeros need not have an accumulation point in the plane. For example,are distinct entire functions agreeing at all the indicated positive integers.