Given the following function:
$f(\theta) =
\begin{cases}
\cos(\theta) + \omega\sin(\theta), & \text{if $0\le\theta<\frac{\pi}{2}$} \\
\omega\cos(\theta-\frac{\pi}{2}) + \omega^2\sin(\theta-\frac{\pi}{2}), & \text{if $\frac{\pi}{2}\le\theta<\pi$} \\
\omega^2\cos(\theta-\pi) + \sin(\theta-\pi), & \text{if $\pi\le\theta<\frac{3\pi}{2}$}
\end{cases}$
where $\omega=-\frac{1}{2} + i\frac{\sqrt{3}}{2}$, is the graph of the function a Reuleaux triangle? If so, how does one prove it?
I tried graphing it and it sure looks like one:
Edit: A huge apology to everyone. The original function I wrote had a gigantic mistake because I misinterpreted the output of the program I wrote. I've already fixed it above. As you can see, the actual function I graphed has a weird domain of just $0\le\theta\lt3\pi/2$, which is probably why I got confused.
Either way, it turns out that dxiv's conclusion is still correct, so I accepted it.
Best Answer
Close, but it is not a Reuleaux triangle. By direct calculation:
$$ \begin{align} f(0) &= 1 \\ f(\pi/2) &= \frac{-1 + i \sqrt{3}}{2} \\ f(3 \pi / 4) &= - \frac{\sqrt{2}}{2} \end{align} $$
Then $\,\left|f(\pi/2) - f(0)\right|^2=3\,$ but $\,\left|f(3 \pi / 4) - f(0)\right|^2 \approx 2.91 \ne 3\,$ so the width is not constant.
[ EDIT ] Updated to match the modification in the question.