I try to reformulate the commentaries of fgp as an answer.
First consider a probability space $(\Omega, \mathcal{A}, \mathbb{P}')$ and a random variable uniformly distributed on $[0,2\pi]$:
$$X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow ([0,2\pi],\mathcal{B}([0,2\pi])).$$
Furthermore consider the parametrization of the unit circle
$$p: [0,2\pi] \rightarrow S^1; \quad x \mapsto e^{i\cdot x}$$
which is continuous.
Now consider the space $(S^1,\mathcal{B}(S^1))$ where $\mathcal{B}(S^1)$ is the $\sigma$-algebra generated by the open sets of $S^1$.
We define $\mathbb{P}$ to be the probability measure induced by the map
$$p\circ X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow (S^1,\mathcal{B}(S^1)); \quad \omega \mapsto e^{i X(\omega)}.$$
Then we have
\begin{equation}
P(Z\in A) = \frac{1}{2\pi i}\int_{A}{\frac{1}{z}}dz \qquad \forall A \in \mathcal{B}(S^1)
\end{equation}
or more generally
$$E[f(Z)]=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz$$
for any measurable, bounded function $f: S^1 \rightarrow \mathbb{R}$
So we must be careful what the measurable space really is. In the case of the "density" in the question we are considering the probability space $(S^1,\mathcal{B}(S^1), \mathbb{P})$ and on this we get can give the value of $\mathbb{P}$ via the above formula.
Hence the above is not a density with respect to the Lebesgue-measure on $\mathbb{C}$, but a way of formulating the value with the help of the parametrization of (or - to be more precise - a contour integral along) the unit circle $S^1$.
The area of the disk is $\pi$. We have a uniform distribution on the disk, so the probability of landing in a part of the disk with area $A$ is proportional to $A$, say $kA$. Since the probability of landing in the unit disk is $1$, we have $k\pi=1$ and therefore $k=\dfrac{1}{\pi}$. Thus the probability of landing in a part of the disk with area $A$ is $\dfrac{A}{\pi}$.
Now what function $f(x,y)$ is it that integrated over any part of the disk with area $A$ gives result $\dfrac{A}{\pi}$? The constant function $\dfrac{1}{\pi}$.
Best Answer
If you generate the points uniformly on the unit circle, you will get the same distribution as if you generate them uniformly on the upper half circle. The fraction of points in the interval $[a,b]$ will be the fraction of the arc length between $x=a$ and $x=b$. The angle at $x=a$ is $\arccos a$ and at $x=b$ is $\arccos b$, so the fraction in $[a,b]$ is $\frac 1{\pi}|\arccos a - \arccos b|$