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

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.

The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.

On the slit domain $D$, choose the complex argument $\theta ={\mathrm{Arg}}_{(0,2\pi )}z$. The resulting holomorphic logarithm is $L(z)=\log |z|+i\theta$, with $L^{\prime }(z)=1/z$. DefineIt is holomorphic as a composition of holomorphic functions, and $f(z{)}^2=e^{L(z)}=z$. At $z=-1$ the chosen argument is $\pi$, so $f(-1)=i$.

Suppose a holomorphic square root $g$ existed on the whole punctured complex plane. It would be nonzero, and differentiating $g^2=z$ would give $2g{g}^{\prime }=1$. There are several ways to express the resulting obstruction; a direct one uses the unit circle. SetThe chain rule and $g^{\prime }(z)=g(z)/(2z)$ imply $q^{\prime }(t)=0$, so $q$ is constant. ButThese cannot agree because $g(1{)}^2=1$. Thus no such global holomorphic square root exists. Going once around zero changes the sign of the locally chosen root.