I'm trying to derive the radial acceleration formula for uniform circular motion: $a =v^2/R$.
Here's what I've $\mathbf r(t)= Re^{-it/2}\mathbf e_1e^{it/2}$ $\implies \ddot {\mathbf r}(t) =\frac R{16} (-\cos(t/2) +i\sin(t/2))\mathbf e_1(-\cos(t/2)-i\sin(t/2)) = \frac 1{16} \mathbf r(t)$
This isn't exactly what I want. What am I doing wrong here?
The easiest way to do this is to use $z(t) = re^{i\omega t}$ to indicate the position vector of the particle as a complex number on the plane of motion. $\omega$ is the angular velocity, and it is constant, as is $r$.
You now get $v(t) = \frac{dz}{dt} = i\omega re^{i\omega t} = i\omega z(t)$, which has magnitude given by $|v(t)| = r\omega$, as expected. The complex form tells you the angular velocity is perpendicular to the position vector, also as expected (remember that multiplying by $i$ rotates a complex number counterclockwise through $\frac{\pi}{2}$.
By differentiating again, you get $a(t) = \frac{dv}{dt} = -\omega^2 re^{i\omega t} = -\omega^2 z(t)$. The magnitude is given by $|a(t)| = r\omega^2$, again as expected. The complex form tells you the acceleration vector is antiparallel to the position vector (i.e. collinear but oppositely directed - that's centripetal acceleration).
You can use the magnitude formulas to rearrange to get the form you require.
$|v(t)| = r\omega \implies \omega = \frac{|v(t)|}{r}$
$|a(t)| = r\omega^2 = r \cdot {(\frac{|v(t)|}{r}})^2 = \frac{{(|v(t)|)}^2}{r}$