Si $f$ y $1/f$ son armónicas $f$ es holomorfa o antiholomorphic

Question

Tengo este problema.

Que $f:D\to \mathbb{C}$ es una función tal que $f$ y $1/f$ son armónicas (las partes real e imaginarias son armónico). Entonces $f$ es holomorfa o antiholomorphic.

He intentado resolver por computación el Laplaciano de partes real e imaginarias, pero resulta muy engorroso. ¿Hay una manera mejor?

Answer 1

Idea posible. Deje $f=f(x,y)$. Si $f$ $1/f$ son armónicas, a continuación, $$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 $$ $$ \frac{\partial^2 1/f}{\partial x^2} + \frac{\partial^2 1/f}{\partial y^2} = 0 $$ donde $$ \frac{\partial 1/f}{\partial x} = -\frac{\partial f}{\partial x} \frac{1}{f^2} \qquad \frac{\partial^2 1/f}{\partial x^2} = -\frac{\partial^2 f}{\partial x^2} \frac{1}{f^2} +2\left(\frac{\partial f}{\partial x}\right)^2\frac{1}{f^3} $$

$$ \frac{\partial 1/f}{\partial y} = -\frac{\partial f}{\partial y} \frac{1}{f^2} \qquad \frac{\partial^2 1/f}{\partial y^2} = -\frac{\partial^2 f}{\partial y^2} \frac{1}{f^2} +2\left(\frac{\partial f}{\partial y}\right)^2\frac{1}{f^3} $$ Así $$ -\frac{\partial^2 f}{\partial x^2} \frac{1}{f^2} +2\left(\frac{\partial f}{\partial x}\right)^2\frac{1}{f^3}-\frac{\partial^2 f}{\partial y^2} \frac{1}{f^2} +2\left(\frac{\partial f}{\partial y}\right)^2\frac{1}{f^3} = \frac{2}{f^3}\left(\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f} {\partial y}\right)^2\right) $$

$$ \frac{2}{f^3}\left(\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f} {\partial y}\right)^2\right)=0 $$ Desde $f \not \equiv 0$, tenemos $$ 0=\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f} {\partial y}\right)^2 =\left(\frac{\partial f}{\partial x}+\frac{\partial f} {\partial y}i\right)\left(\frac{\partial f}{\partial x}-\frac{\partial f} {\partial y}i\right) $$

Answer 2

Que $\frac{\mathrm{d}}{\mathrm{d}z}=\tfrac12(\frac{\mathrm{d}}{\mathrm{d}x}-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}y})$ y $\frac{\mathrm{d}}{\mathrm{d}\overline{z}}=\tfrac12(\frac{\mathrm{d}}{\mathrm{d}x}+\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}y})$ como de costumbre. Entonces $$\frac{\mathrm{d}^2}{\mathrm{d}z \, \mathrm{d}\overline{z}} = \frac14\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}+ \frac{\mathrm{d}^2}{\mathrm{d}y^2}\right)$$ and $f$ is holomorphic if $\frac{\mathrm{d}f}{\mathrm{d}\overline{z}}=0$ or anti-holomorphic if $\frac{\mathrm{d}f}{\mathrm{d}z}=0$. Now under the given conditions $$0 = \frac{\mathrm{d}^2 f^{-1}}{\mathrm{d}z \, \mathrm{d}\overline{z}} = 2 f^{-3} \frac{\mathrm{d}f}{\mathrm{d}z} \frac{\mathrm{d}f}{\mathrm{d}\overline{z}}.$$