= Solution
At a zero $a$, <Taylor's theorem> gives $f(z)=\sum_{n\geq0}c_n(z-a)^n$ near $a$. If some <coefficient> is nonzero, let $m$ be the least such index. Then $f(z)=(z-a)^mq(z)$ with $q(a)\ne0$, and <continuity> makes $q$ nonzero on a smaller disc. Thus $a$ is an isolated zero. A <limit point> of zeros is also a zero by <continuity> and cannot be isolated, so all Taylor <coefficients> there vanish and $f$ vanishes on a nonempty open disc.
Let $A$ be the set of points having a neighborhood on which $f$ vanishes. It is nonempty and open. A point in its relative closure either already belongs to $A$ or is a <limit point> of distinct zeros; the preceding argument puts it in $A$ in either case. Thus $A$ is also relatively closed. <Connectedness> of $D$ gives $A=D$, proving \b[$f$ is identically zero].
For conjugate reflection, the disc is invariant under conjugation, and
$$
\frac{h(z+w)-h(z)}w=\overline{\frac{g(\bar z+\bar w)-g(\bar z)}{\bar w}}\longrightarrow\overline{g'(\bar z)}.
$$
Hence \b[$h$ is <holomorphic>], with <derivative> $\overline{g'(\bar z)}$. If $g$ is real on the real diameter, then for real $x\in(-1,1)$, $h(x)=\overline{g(x)}=g(x)$. The <holomorphic> difference $g-h$ therefore has zeros accumulating at zero, and the <identity theorem> just proved gives \b[$g=h$ on the entire disc].
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