A Green's function is defined as follows:
$$G(\omega) = \frac{1}{N}\mathrm{E}\big[ \mathrm{Tr}\frac{1}{I\omega - J} \big]$$, where $I$ is the $N$-dimensional identity and $E$ means expectation value with respect to the random matrix $J$.


It follows that 
$$ G(\omega) = \frac{1}{N}\mathrm{E}\big[ \sum\_\lambda\frac{1}{\omega - \lambda} \big] = \int d^2\lambda\frac{\rho(\lambda)}{\omega - \lambda},$$
where $\rho(\lambda)$ is the average density of eigenvalues $\lambda$ of $J$ in the complex plane.


I am confused as to how this equality is obtained. Can someone help to explain or present the idea behind it ?



Alternative formula of a Green's function for average density of eigenvalues of random matrix
I'll post an answer to spell out all the details.


You have $$G(ω)=\frac{1}{N}E\left[{\rm Tr}\frac{1}{Iω−J}\right]=\frac{1}{N}E\left[\sum\_\lambda\frac{1}{ω−\lambda}\right]$$


This can be written as $$G(ω)=\frac{1}{N}E\left[\int d^2z \sum\_\lambda\frac{\delta(z-\lambda)}{ω−z}\right],$$ where $\delta$ is the delta-function/distribution in the complex plane.


Now define $$\rho(z)=\frac{1}{N}E\left[\sum\_\lambda\delta(z-\lambda)\right]$$, which is the spectral density. Then it follows that $$G(\omega)=\int d^2z \frac{\rho(z)}{\omega-z}=$$


Or maybe you mean it the other way around. If you start with $$\frac{1}{\omega+i\epsilon-\lambda}=\frac{\omega-i\epsilon-\lambda}{(\omega-\lambda)^2+\epsilon^2}$$, the imaginary part gives, in the limit of vanishing $\epsilon$ (as Beenakker mentioned), $$\lim\_{\epsilon\to 0}{\rm Im}E\left[\sum\_\lambda\frac{1}{\omega+i\epsilon-\lambda}\right]=\frac{1}{\pi}E\left[\sum\_\lambda\delta(\omega-\lambda)\right]$$,
which means that $\lim\_{\epsilon\to 0}{\rm Im}G(\omega+i\epsilon)=\rho(\omega)N/\pi$.


