While this is really more of a non-answer to your question (I'm not really fond of such philosophical generalities), I think part of what sets complex geometry apart is the *lack* of a Frobenius.

If one wants to make use of the Frobenius in studying complex geometry, it is still possible to do so by reducing to characteristic $p$. For instance, the simplest proof of the next result uses this method (the following discussion is a summary from the beautiful book "Higher-Dimensional Algebraic Geometry" by Debarre).

**Theorem.** If $X$ is a smooth Fano variety (i.e. $-K_X$ is ample) over $\mathbb{C}$, then through any point of $X$ there is a rational curve.

It is relatively easy to prove this result in characteristic $p$ by making use of the Frobenius. For an outline, first fix a point $x\in X$. If you can produce a map $f:C\to X$ from a pointed curve $(C,c)$ taking $c$ to $x$ in such a way that this map can be deformed while holding $f(c)=x$ constant, then by Mori's bend-and-break lemma it is possible to find a rational curve through $c$. Now we can estimate the dimension of the space of such deformations, and it follows that if $$-K_X \cdot f_\ast C - n g(C) \geq 1 \qquad (n=\dim X)$$ that we will be done.

Now the question becomes: how can we produce a curve $C$ (or rather a map $f$) such that this inequality holds? If you play around in complex geometry, you will find that standard constructions, such as taking branched covers of a given curve, won't get you anywhere: if you manage to increase the $(-K_X)$-degree $-K_X\cdot f_\ast C$, then this will lead to an increase in the genus $g(C)$ as well, and you won't have made any progress towards proving the inequality.

However, in characteristic $p$, it is possible to increase $-K_X\cdot f_\ast C$ while keeping the genus of $C$ fixed! If we start from any random map $f:C\to X$ with $f(c)=x$, we can precompose it as many times as we wish by the Frobenius $Frob:C\to C.$ This has the effect of multiplying $-K_X\cdot f_\ast C$ by $p$, while the genus $g(C)$ remains constant since the domain curve is the same (although different as a $k$-scheme). Then since $-K_X$ was ample, it follows that if we compose with the Frobenius enough times our inequality will be satisfied.

Proving that the characteristic $p$ case implies the characteristic $0$ case is a relatively formal argument; I'll refer the reader to Debarre to see it.