= Solution
{c}
Let $F(z)=\overline{\sinh z}$. More generally, for a <holomorphic function> $h$, the increment expansion at $z_0$ gives
$$
\frac{\overline{h(z_0+w)}-\overline{h(z_0)}}w=\overline{h'(z_0)}\frac{\overline w}w+o(1).
$$
Along real increments $w$ the first factor $\overline w/w$ is $1$, while along purely imaginary increments it is $-1$. Thus <complex differentiability> of $\overline h$ requires $h'(z_0)=0$. Conversely, if $h'(z_0)=0$, the displayed expression tends to zero for arbitrary complex increments, so the <complex derivative> exists and equals zero.
For $h=\sinh$, this condition is $\cosh z_0=0$, equivalent to $e^{2z_0}=-1$. Consequently
$$
\boxed{F\text{ is complex differentiable exactly at }z_0=(k+\tfrac12)\pi i,\quad F'(z_0)=0.}
$$
These points form a <discrete subset> of the <complex plane> and contain no nonempty <open set>. Every <neighbourhood> of each one contains points where $F$ is not <complex differentiable>. Therefore $F$ is <holomorphic> on no nonempty <open set>, even though it has <complex derivatives> at the specified points.
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