Let $L$ be the left shift operator on $\ell^2(\mathbb{Z})$ with trace $\tau(T) := \langle T \delta_0, \delta_0 \rangle$.

How can I show that the Brown measure of $L$ is the uniform measure on the unit circle?

The Brown measure is defined as follows: For each $z \in \mathbb{C}$ define $\nu_{z}$ to be the spectral measure of the (self-adjoint) operator $(L - z)^*(L-z)$. Then let $$f(z) = \frac{1}{2} \int_0^\infty \log x \,d\nu_z(x).$$ Then the Brown measure is defined as $$\mu_L := \frac{1}{2\pi} \Delta f,$$ where the Laplacian is taken in the sense of distributions. Motivation for the definition can be found at Terence Tao's notes on the circular law for random matrices.

Edit: posted at math stackexchange https://math.stackexchange.com/questions/2258115/brown-measure-of-left-shift-operator