So far I have rewritten the following expression:
$$f'(z) = \lim_{\Delta z\rightarrow0}\frac{f(z + \Delta{z})-f(z)}{\Delta{z}}$$
As such:
$$f'(z) =\lim_{\Delta z\rightarrow 0} (\bar{z} + \bar{\Delta{z}} + z\frac{\bar{\Delta{z}}}{\Delta{z}}) $$
This limit only exists if each of the three terms in the summation have limits. Clearly the first two do, so I am currently evaluating the third term, that is, seeing if.. $$z(\lim_{\Delta{z}\rightarrow 0}\frac{\bar{\Delta{z}}}{\Delta{z}})$$
..exists for $z=0$. I don't think the limit above exists: if we allow $\Delta{z}$ to approach $0$ through the real numbers, then the numerator and denominator are the same and the limit evaluates to 1. Else, if we let $\Delta{z}$ approach $0$ from 'above' (through the imaginary numbers), we have the denominator is the negation of the numerator and so the limit is $-1$. Hence the limit is undefined.
However, would $0$ times a limit that does not exist be $0$? I'm just not sure how to prove this term does indeed evaluate to some value instead of being undefined; it's clear that for $z \neq 0$ that the expression is undefined, but for $z=0$ I'm confused if $0$ times an undefined limit is $0$?