Consider the function:
$$y = \frac{1}{3}x^3 + x$$
Suppose we wanted to determine its local optima, but instead of looking at local optima with domain $R$ we instead consider domain $C$ and range $C$. We thus continue the process as is intuitive:
$$y' = x^2 + 1$$ $$y'=0 \rightarrow x = \pm i $$
Now clearly the two stationary points for the second derivative are at the complex values of $\pm i$. The normal intuition of them being relatively less than their surroundings and relatively more than their surroundings doesn't appear to make immediate sense since complex numbers aren't directly comparable.
So I came up with the following hypothesis
Given
$$y = f(x)$$ All points in C where $$y' = f'(x) = 0$$ Are local optima for $$y = |f(x)|$$
Is this correct? How do I show it?
I would be perplexed if it weren't.