Can we conclude that $u^{-1}+iu$ is constant?

Question

If $u$ is a real valued function on $\Delta _R$ and $u^{-1}+iu$ is analytic in $\Delta_R$. Then can we conclude that $u$ is constant?

Answer 1

I'm assuming $\Delta_{\Bbb R^2}$ is an open set in $\Bbb C$. I also assume that $u \ne 0$ in $\Delta _{\Bbb R^2}$, so that $u^{-1}$ is well-defined there. I further assume that "analytic" means "complex analytic"; this based on the "complex analysis" tag.

Soooo, if $u^{-1} + iu$ is analytic, then both $u$ and $u^{-1}$ are harmonic, whence

$\nabla^2 u = \nabla^2 u^{-1} = 0. \tag{1}$

We compute $\nabla^2 u^{-1}$:

$\nabla^2 u^{-1} = \nabla \cdot \nabla u^{-1}; \tag{2}$

$\nabla u^{-1} = -u^{-2} \nabla u; \tag{3}$

$\nabla \cdot (-u^{-2} \nabla u) = 2u^{-3} \nabla u \cdot \nabla u - u^{-2} \nabla \cdot \nabla u = 2u^{-3} \vert \nabla u \vert^2 - u^{-2} \nabla^2 u = 2u^{-3} \vert \nabla u \vert^2, \tag{4}$

using(1). Thus (1) also implies

$2u^{-3} \vert \nabla u \vert^2 = 0. \tag{5}$

Now we see that $u \ne 0$ forces

$\nabla u = 0: \tag{6}$

thus $u$ must be constant on each connected component of $\Delta_{\Bbb R^2}$. QED.

Hope this helps. Cheerio,

and as ever,

Fiat Lux!!!