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I have this problem.

Let $f:D\to \mathbb{C}$ be a function such that $f$ and $1/f$ are harmonic (Their real and imaginary parts are harmonic). Then $f$ is holomorphic or antiholomorphic.

I tried to solve it by computing the laplacian of real and imaginary parts, but it becomes very cumbersome. Is there a better way?

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    $\begingroup$ It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $\frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z}f=0$. Then, compute $0=\frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z}(\frac{1}{z}\circ f)$. The chain rule gives you $=\frac{\partial}{\partial \overline{z}}(-\frac{1}{f^2})\cdot \frac{\partial f}{\partial z}+\left(-\frac{1}{f^2}\right)\cdot \frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z}f$ ... $\endgroup$ – user577471 Jul 20 '18 at 17:18
  • $\begingroup$ So, either $\frac{\partial f}{\partial z}=0$, in which case $f$ is anti-holomorphic, or $\frac{\partial}{\partial\overline{z}}\left(-\frac{1}{f^2}\right)=0$. Do another chain rule to get $2\cdot \frac{1}{f^3}\frac{\partial f}{\partial\overline{z}}=0$, which gives you that $f$ is holomorphic. $\endgroup$ – user577471 Jul 20 '18 at 17:21
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Possible idea. Let $f=f(x,y)$. If $f$ and $1/f$ are harmonic then $$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 $$ $$ \frac{\partial^2 1/f}{\partial x^2} + \frac{\partial^2 1/f}{\partial y^2} = 0 $$ where $$ \frac{\partial 1/f}{\partial x} = -\frac{\partial f}{\partial x} \frac{1}{f^2} \qquad \frac{\partial^2 1/f}{\partial x^2} = -\frac{\partial^2 f}{\partial x^2} \frac{1}{f^2} +2\left(\frac{\partial f}{\partial x}\right)^2\frac{1}{f^3} $$

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

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

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Let $\frac{\mathrm{d}}{\mathrm{d}z}=\tfrac12(\frac{\mathrm{d}}{\mathrm{d}x}-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}y})$ and $\frac{\mathrm{d}}{\mathrm{d}\overline{z}}=\tfrac12(\frac{\mathrm{d}}{\mathrm{d}x}+\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}y})$ as usual. Then $$\frac{\mathrm{d}^2}{\mathrm{d}z \, \mathrm{d}\overline{z}} = \frac14\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}+ \frac{\mathrm{d}^2}{\mathrm{d}y^2}\right)$$ 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 $$0 = \frac{\mathrm{d}^2 f^{-1}}{\mathrm{d}z \, \mathrm{d}\overline{z}} = 2 f^{-3} \frac{\mathrm{d}f}{\mathrm{d}z} \frac{\mathrm{d}f}{\mathrm{d}\overline{z}}.$$

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