 Solution

The identity theorem can be expressed as the principle of isolated zeros: a holomorphic function on a connected open set is either identically zero or has isolated zeros. Locally this follows from its Taylor series: at a zero of a function which is not locally zero, factor the first nonzero power, $f(z)=(z-z_0{)}^{m}h(z)$ with $h(z_0)\ne 0$. Connectedness then rules out a nonempty open region of zeros unless the whole function vanishes.

Set $g(z)=\overline{f(\bar{z})}$. This is holomorphic on the punctured plane: reflecting and conjugating a local power series produces another power series. The absence of an essential singularity means that the singularity at zero is removable or a pole. Thus for some integer $m\ge 0$, $z^{m}f(z)$ extends holomorphically across zero; the same is true of $z^{m}g(z)$. Their differencetherefore extends to a holomorphic function on all of $\mathbb{C}$. At every $1/n$, real-valuedness gives $H(1/n)=0$. These zeros accumulate at zero, which is now an interior point of the domain of $H$. The identity theorem forces $H\equiv 0$. For $z\ne 0$ we may divide by $z^m$, yieldingThe conjugation outside $f$ is essential. Also, applying the identity theorem directly on the punctured plane would be invalid: the accumulation point zero does not belong to that domain.