Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 3 / a

The hyperbolic sine satisfies $\sinh z=(e^z-e^{-z})/2$. The complex exponential function is entire, and sums and compositions of entire functions are entire. Therefore

The zeros of the hyperbolic sine are exactly $z=k\pi i$, $k\in \mathbb{Z}$, since $\sinh z=0$ is equivalent to $e^{2z}=1$. Each zero is simple because $\cosh (k\pi i)=(-1{)}^k\ne 0$. The quotient is therefore holomorphic precisely onAt zero, the Taylor expansion $\sinh z=z+O(z^3)$ gives $z/{\sinh }^{2}z=1/z+O(z)$, so zero is a simple pole. At $k\pi i\ne 0$, the denominator has a double zero and the numerator is nonzero, giving a pole of order two. Thus none of the omitted points admits a holomorphic extension.

The denominator vanishes exactly when $e^z=-1$, namely at $z=(2k+1)\pi i$, $k\in \mathbb{Z}$. Its derivative there is $e^z=-1\ne 0$, so each such zero is simple and the numerator does not cancel it. Henceand every omitted point is a simple pole.

Let $F(z)=\overline{\sinh z}$. More generally, for a holomorphic function $h$, the increment expansion at $z_0$ givesAlong 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 $h^{\prime }(z_0)=0$. Conversely, if $h^{\prime }(z_0)=0$, 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 $e^{2z_0}=-1$. ConsequentlyThese 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.