Modulus and argument of a holomorphic function. It's well known that $\Re(z)$ and $\Im(z)$ are harmonic functions (where $z$ is a holomorphic function). What about the modulus and the argument?
 A: No and no. 
For the modulus, note that if $f(z)=z^2$, then $\Delta |z|^2 = 4 \neq 0$.
For the argument, the situation is even worse: there is no continuous choice of argument for $f$ (that is, there is no continuous function $\theta : \mathbb C \to \mathbb R$ such that $f(z)=|z|^2 e^{i \theta(z)}$. Since harmonic functions are continuous, the argument cannot be harmonic.
A: With the reservation stated in the other answer, the different branches of the argument are harmonic in subsets strictly lesser than the plane. For example:
$$\arg(x + yi) = \arctan(y/x)\hbox{ (main branch of $\arctan$)}$$
is harmonic:
$$
\frac{\partial^2}{\partial x^2}\arctan(y/x) +
\frac{\partial^2}{\partial y^2}\arctan(y/x) =
\frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} =0.
$$
Modulus isn't harmonic, but its $\log$ is harmonic in $\Bbb C\setminus\{0\}$:
$$\frac{\partial^2}{\partial x^2}\log\left(\sqrt{x^2 + y^2}\right) +
\frac{\partial^2}{\partial y^2}\log\left(\sqrt{x^2 + y^2}\right) =
\frac12\left(\frac{y^2 - x^2}{(x^2 + y^2)^2}\right) +
\left(\frac{x^2 - y^2}{(x^2 + y^2)^2}\right) = 0.
$$
