Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 3

At every $z=x+iy\in U$, the Cauchy-Riemann equations areTo derive them from complex differentiability, approach $z$ along real increments $h$. The complex derivative givesApproaching along imaginary increments $ih$ instead givesThese limits exist because $f$ is holomorphic. Equating their real and imaginary parts proves the equations, together with $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$.

A sufficient condition is that $U$ is a connected space. Suppose, towards a contradiction, that $|f|$ is the same constant $R$ throughout $U$. If $R=0$, then $f=0$, contradicting $f(a)\ne f(b)$.

If $R>0$, differentiate $u^2+v^2=R^2$ in $x$ and $y$, and use the Cauchy-Riemann equations from part (a). They giveThe coefficient matrix for $(u_x,v_x)$ has determinant $-(u^2+v^2)=-R^2\ne 0$. Hence $u_x=v_x=0$, so $f^{\prime }=0$ everywhere. A holomorphic function with zero complex derivative is locally constant: within a small disc, differentiate along a straight segment. Local constancy and connectedness imply constancy on $U$, again a contradiction.

Consequently,This is the constant-modulus holomorphic function principle. The extra hypothesis matters: on two disjoint open discs, define $f=1$ on one and $f=-1$ on the other. It is holomorphic and nonconstant, yet $|f|=1$ everywhere.

The function $h(z)=1/(z+1)$ is holomorphic on $U$ and hasIts imaginary part is exactly the prescribed $v$. Thus $f-h$ is a real-valued holomorphic function. The Cauchy-Riemann equations force both partial derivatives of its real part to vanish. Since $\mathbb{C}\setminus \{-1\}$ is a connected space, $f-h$ is a single real constant. All possibilities are thereforeConversely every such function has the stated imaginary part. The data determine $f$ only up to this real additive constant.

The identity theorem states: if $V\subseteq \mathbb{C}$ is a nonempty connected open set and $p,q$ are holomorphic functions on $V$, and their equality set has a limit point in $V$, thenThe interior location of the limit point and connectedness are essential hypotheses.

For the proof put $h=p-q$. At any $z_0\in V$, its convergent Taylor series either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\ge 0$. In the latter case,where $k$ is holomorphic near $z_0$. By continuity, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior limit point therefore forces the neighbourhood alternative.

Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne \varnothing$, and $E$ is an open set. It is also a closed set relative to $V$: a point of its closure outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same Taylor series argument would place it in $E$. Since $V$ is connected, $E=V$. Thus $h=0$ on $V$, proving the theorem.

Define $F(z)=\overline{f(\overline{z})}$. To apply the identity theorem, we first verify that $F$ is holomorphic; mere complex conjugation of the values of $f$ would not suffice. For $h\ne 0$,Thus $F$ is an entire function, just as $f$ is. At each positive integer $n$, the point $1/n$ is real and $f(1/n)$ is real, so $F(1/n)=f(1/n)$. These distinct points accumulate at $0\in \mathbb{C}$. The identity theorem on the connected complex plane givesThis is conjugate reflection of a holomorphic function, with the reality data forcing equality to its reflection.