= 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>, define
$$
S=\{z\in D:h^{(n)}(z)=0\text{ for every }n\geq0\}.
$$
The 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 \b[$f=g$ throughout $D$], proving the <identity theorem for holomorphic functions> rather than merely invoking it.
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