 Solution

At a zero $a$, Taylor's theorem gives $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ near $a$. If some coefficient is nonzero, let $m$ be the least such index. Then $f(z)=(z-a{)}^{m}q(z)$ with $q(a)\ne 0$, 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 $f$ is identically zero.

For conjugate reflection, the disc is invariant under conjugation, andHence $h$ is holomorphic, with derivative $\overline{g^{\prime }(\bar{z})}$. 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 $g=h$ on the entire disc.