If $f$ and $1/f$ are harmonic then $f$ is holomorphic or antiholomorphic 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?
 A: 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
$$
0=\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f} {\partial y}\right)^2 =\left(\frac{\partial f}{\partial x}+\frac{\partial f} {\partial y}i\right)\left(\frac{\partial f}{\partial x}-\frac{\partial f} {\partial y}i\right)
$$
A: 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 $$0 = \frac{\mathrm{d}^2 f^{-1}}{\mathrm{d}z \, \mathrm{d}\overline{z}} = 2 f^{-3} \frac{\mathrm{d}f}{\mathrm{d}z} \frac{\mathrm{d}f}{\mathrm{d}\overline{z}}.$$
