How do I *show* that the derivative of this simple complex function exists? Consider this simple (complex) function:
$$f(z)=z^{3/2}$$ and I want to see for what values of $z$ does this have a derivative.
Since I know that the derivative rules for complex functions are the same as those of real functions is can simply find the derivative: $$f'(z)=\frac{3}{2}z^{1/2}$$
So it has a derivative in the whole of the argand plane.
But I have issues in trying to prove it:
I need to essentially show that the following limit is independent of the path taken in:  $$f'(z)=Lim_{\Delta z\to 0}\frac{(z+\Delta z)^{3/2}-z^{3/2}}{\Delta z}$$
With $3/2$ power it is not clear how I can evaluate this limit. I could do a Binomial expansion but it seems there must be an easier to prove that the derivative exists without having to actually calculate this limit this way.
 A: First of all, you need to be clear what you mean by $z^{3/2}$. Explicitly, for each $z=re^{i\theta}$, $r\ge0$, $\theta\in[0,2\pi)$,  define,
$$f(z)=r^{3/2}e^{i3\theta/2}$$
Notice, that I picked a "branch" to use for my definition. Meaning I chose an interval $[0,2\pi)$. I could have chosen any interval (or some other simply connected region). Hence there are many many definitions of $z^{3/2}$ that are all valid. Simply writing $z^{3/2}$ as though you are not talking about a family of possible functions is in error.
We note immediately that $f$ fails to be continuous at the cut I made, in this case it is not continuous on $[0,\infty)$. This is because approaches from quadrant I and IV do not have the same limit. This means that $z^{3/2}$ is not differentiable on the cut either.
You are correct, that where ever the function is differentiable, the derivative will be given by
$$f'(z)=\frac{3}{2}z^{1/2}$$
Assuming we are discussing the same branch cuts for both functions. In general, whenever someone writes $z^{3/2}$ they automatically mean "with respect to some branch cut".
Now if you want to go about proving this using a limit, then the classical approach is to simply use the root-conjugate.
$$\lim_{\Delta z\to 0} \frac{(z+\Delta z)^{3/2}-z^{3/2}}{\Delta z}\cdot \frac{(z+\Delta z)^{3/2}+z^{3/2}}{(z+\Delta z)^{3/2}+z^{3/2}}=$$
$$\lim_{\Delta z\to 0} \frac{3z^2\Delta z+3z\Delta z^2+\Delta z^3}{\Delta z ((z+\Delta z)^{3/2}+z^{3/2})} = \frac{3z^2}{2z^{3/2}} $$
But again, all of those calculations are under the assumption that we are restricted to some branch of the root function.
