Complex rates of change?

Question

For the function $f(x)=e^{ix}$ defined over $\mathbb{C}$, if its derivative is $ie^{ix}$ does that mean rates of change at various points are complex numbers? And what does that suggest about the nature of the function?

This was just a thought I came up with. It doesn't make much sense to think about directly but I was thinking it might mean something. Like for example the rate of change at $0 \implies i$.

Answer 1

Yes it does. It goes back to the definition of a derivative. $$f'(z)=\lim_{h \to 0}\frac {f(z+h)-f(z)}h$$ When you work in the complex plane $h$ is allowed to be complex. It still means that if you change the input by $h$ the function value changes by $hf'(z)$. In your example, if $f(z)=e^{iz}, f'(z)=ie^{iz}$ we have $f(0)=1, f(h)=e^{ih}$. For $h$ small, $e^{ih}\approx 1+ih$ so the change in the function value, $ih$ is $i$ times the change in input, $h$.

Answer 2

Rather than "rate of change" think of it as "velocity". In two dimensions, velocity is a vector. Here, that vector is considered to be a complex number.