Can a $2D$ canonical transformation also be holomorphic?

Question

In Hamiltonian mechanics we have canonical transformations on an even-dimensional space. What if we consider a canonical transformation of two variables $f(x,y)=u(x,y)\mathbf{e_1}+v(x,y)\mathbf{e_2}$ as a function on $\mathbb{C}$ and require that the transformation is also holomorphic? What one gets by combining the Cauchy-Riemann equations with the symplectic condition $\{u,v\}=1$ is that \begin{equation} (\partial_xu)^2+(\partial_yu)^2=1 \end{equation} and the same applies to $v$.

Question: What are some examples of a such function $f$?

I can think of a component e.g. $u$ as \begin{equation} u(x,y) = \frac{x}{\sqrt{x^2+y^2}} \end{equation} that satisfies the condition above but then Iam not anble to find a $v$ such that $f$ is holomorphic.
I tried with something as \begin{equation} u(x,y) = \sin\theta(x,y)\qquad v(x,y)=\cos\theta(x,y) \end{equation} but no luck. Any ideas?

Answer 1

tl; dr: Any such mapping is a holomorphic Euclidean motion, i.e., of the form $f(z) = e^{i\theta}z + b$ for some real $\theta$ and some complex $b$.

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Let $(x, y)$ denote Cartesian coordinates in $D$, and write $z = x + iy$.

Lemma Let $D$ be a plane region (connected, non-empty open set). If $g$ is holomorphic in $D$ and $|g|^{2} = 1$ in $D$, then $g$ is constant.

Proof: Standard complex analysis exercise, see spoiler.

Write $g = s + it$ with $s$ and $t$ real-valued. By hypothesis, $g$ is holomorphic and $|g|^{2} = s^{2} + t^{2} = 1$ in $D$. Differentiating with respect to $x$ and cancelling the factor of $2$ gives $ss_{x} + tt_{x} = 0$. Differentiating with respect to $y$, cancelling $2$, and using Cauchy-Riemann gives $ss_{y} + tt_{y} = -st_{x} + ts_{x} = 0$. Squaring these equations and adding gives, after a bit of algebra, $(s^{2} + t^{2})(s_{x}^{2} + t_{x}^{2}) = 0$. Since $s^{2} + t^{2} = 1$ and $s_{x}$, $t_{x}$ are real-valued, we deduce $s_{x} = t_{x} = s_{y} = t_{y} = 0$ in $D$. By the mean value theorem, $g$ is constant in $D$.

Proposition: Assume $u$ and $v$ are real-valued functions in a plane region $D$, and that $f = u + iv$ is both holomorphic and area-preserving in $D$. Then $f$ is affine, of the form $f(z) = e^{i\theta}x + b$ for some real $\theta$ and some complex $b$.

Proof: Because $f$ is holomorphic, the Cauchy-Riemann equations $u_{x} = v_{y}$ and $u_{y} = -v_{x}$ are satisfied in $D$. Because $f$ is area-preserving, $$ 1 = \det Df = u_{x}v_{y} - u_{y}v_{x} = (u_{x})^{2} + (u_{y})^{2} = |f'|^{2}. $$ Since $f'$ itself is holomorphic, it is constant by the lemma. We therefore have $f'(z) = e^{i\theta}$ for some real $\theta$. The identity theorem implies that for some complex $b$, we have $f(z) = e^{i\theta}z + b$ in $D$.