 Solution

Put $h=f-g$. Choose a disc about $z_0$ contained in the domain $D$. Part (a) gives a convergent Taylor series there, and the accumulating zeros together with part (b) force all its coefficients to vanish. Thus $h=0$ near $z_0$.

To propagate this to the entire domain, defineThe set is nonempty by the preceding local argument. Each derivative is continuous, so $S$ is closed relative to $D$ as the intersection of their zero sets. If $z\in S$, its Taylor series is zero on a disc about $z$, and all derivatives vanish throughout that disc; hence $S$ is open relative to $D$. Connectedness forces $S=D$. Therefore $f=g$ throughout $D$, proving the identity theorem for holomorphic functions rather than merely invoking it.