Why is this? How is $i$ the slope of the function? Where is it the slope?
I understand taking the derivative with the power rule in, for example, the parabola $x^2$ becoming $2x$ and seeing where that is the slope, but I don't understand how dividing two things a number was multiplied by gives you a derivative exactly.
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You have$$f'(x)=\lim_{h\to0}\frac{i(x+h)-ix}h=\lim_{h\to0}i=i.$$But this has nothing to do with the concept of slope. That's from differentiable real functions of one real variable.
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The idea of “slope” is a useful intuition when you are learning single variable (real) calculus, but has more general meanings when dealing in higher dimensions.
In particular, the field of complex numbers has dimension $2$ over the reals, so the graph of a function from the reals to the complex has degree $3.$ So “slope” has to be generalized.
A simpler way to think about this case is to think of $f$ as a function of time, $f(t)=it,$ returning a position at time $t.$ Then at time $t$ the velocity vector, the derivative, is $i.$
The values taken by the function $f : x \in \mathbb{R} \, \longmapsto \, ix$ are complex numbers. However that's not a big deal. You can still define the derivative of $f$ using a limit. Given $x \in \mathbb{R}$,
$$ f'(x) = \lim \limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim \limits_{h \to 0} \frac{i(x+h) - ix}{h} = i. $$