If $a(\Re f(z))^2+b(\Im f(z))^2=1$ $\forall z\in D$ for some $a,b > 0$ and holomorphic$ f$, then $f$ is constant.

Question

Given $D \subset \mathbb{C}$, if $f:D\to\mathbb{C}$ is a holomorphic function such that $a(\Re f(z))^2+b(\Im f(z))^2=1$ $\forall z\in D$ for some $a,b > 0$, then $f$ is constant.

Answer 1

We note that for any functions $u$ and $v$,

$\nabla^2(uv) = \nabla \cdot \nabla(uv) = \nabla \cdot (u\nabla v + v \nabla u) = \nabla \cdot (u \nabla v) + \nabla \cdot (v\nabla u)$ $= \nabla u\cdot \nabla v + u \nabla^2 v + \nabla v \cdot \nabla u + v \nabla^2 u = u\nabla^2 v + v\nabla^2 u + 2\nabla u \cdot \nabla v, \tag 1$

which is a consequence of the easily verified identities

$\nabla(uv) = u \nabla v + v\nabla u \tag 2$

$\nabla \cdot (u \nabla v) = \nabla u \cdot \nabla v + u \nabla^2v; \tag 3$

with $u$ and $v$ harmonic (1) becomes

$\nabla^2(uv) = 2\nabla u \cdot \nabla v; \tag 4$

thus,

$\nabla^2(u^2) = 2\nabla u \cdot \nabla u = 2 \vert \nabla u \vert^2. \tag 5$

Taking

$f = u + iv \tag 6$

we have

$au^2 + bv^2 = a(\Re f)^2 + b(\Im f)^2 = 1, \tag 7$

whence via (5),

$2a\vert \nabla u \vert^2 + 2b\vert \nabla v \vert^2 = a\nabla^2 u^2 + b\nabla^2 v^2 = 0; \tag 8$

it follows then since $a, b > 0$ that

$\vert \nabla u \vert = \vert \nabla v \vert = 0, \tag 9$

hence $u$, $v$ and $f = u + iv$ are all constant. $OE\Delta$.