Can one uniformly generate complex numbers of absolute value less than a given constant R?
This would appear to be equivalent to picking points $(x,y)$ uniformly in a disk of radius R, where $x$ is the real component of the complex number, and $y$, its complex component.
This question appears to differ from numerous others that have been asked, in that $R$ is not assumed to equal 1 (that is, the unit disc).
(Note: I am a Mathematica user MathematicaQuestions, not an R [no pun, intended] user.)
The MathWorld entry DiskPointPicking asserts (without an explicit proof) that to generate uniformly distributed points ($x,y)$ in the unit disk, one should employ \begin{equation} x=\sqrt{r} \cos{\theta},\hspace{.2in} y=\sqrt{r} \sin{\theta}. \end{equation} where $r \in [0,1]$, and $\theta \in [0, 2 \pi]$ are uniformly distributed variables.
So for points ($x,y)$ in a disk of radius $R$, it appears then that one should employ \begin{equation} x=\sqrt{\tilde r} \cos{\theta},\hspace{.2in} y=\sqrt{\tilde r} \sin{\theta}. \end{equation} where $\tilde r \in [0,R^2]$, and $\theta \in [0, 2 \pi]$ are uniformly distributed variables.
Upon further reflection/application, I'm somewhat confused here by my original answer (trying to apply the MathWorld argument).
Say, the maximum absolute value $R$ of the complex numbers $x+ I y$, I want to generate is $\frac{1}{2}$. Now (assuming $x=y$, which seems permissible), we have the relation, \begin{equation} \sqrt{(\frac{1}{\sqrt{8}})^2+(\frac{1}{\sqrt{8}})^2}=\sqrt{\frac{1}{4}}= \frac{1}{2}. \end{equation}
So, it seems that I want to choose $\tilde{r}$--before taking its square root--from $[0,\frac{1}{8}]$, not $[0,R^2 =\frac{1}{4}]$.
???