How to find the set for which a function is harmonic LEt $f(z) = Im( z + \frac{1}{z} ) $. I need to find the set where $f$ is harmonic. IS there a way to find this without too much computations?
 A: The function $g(z) = z + \frac{1}{z}$ is holomorphic everywhere but at $0$, this means that the real and imaginary parts satisfy the Cauchy Riemann equations if $z \neq 0$, this imples that $g$ is harmonic for $z \neq 0$. Hence the real and imaginary parts are harmonic for $z \neq 0.$ So all you need to show is that $f$ is not defined at $z \neq 0$, and you're done!
A: Although @Almentoe has shown that $f$ is harmonic is the punctured plane that deleted the origin, here we analyze the problem directly - "brute force."
First, note that we can write for $z\ne 0$
$$f(z)=\text{Im}\left(z+\frac1z\right)=y-\frac{y}{x^2+y^2}$$  
Then, for $(x,y)\ne (0,0)$, we have
$$\frac{\partial^2 f}{\partial x^2}=\frac{2y(y^2-3x^2)}{(x^2+y^2)^3} \tag 1$$
and
$$\frac{\partial^2 f}{\partial y^2}=-\frac{2y(y^2-3x^2)}{(x^2+y^2)^3} \tag 2$$
Adding $(1)$ and $(2)$ reveals that for $(x,y)\ne (0,0)$, 
$$\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$$
We conclude that $f$ is harmonic for $z\ne 0$.
