Trying to work out if $a^z$ is complex differentiable I am trying to work out if $f\left(z\right)=a^z$ is complex differentiable but I have never actually been taught any complex analysis and I am mostly using wikipedia and the internet in general to do research so I do not know if the approach I am taking is valid. I am trying to use the Cauchy–Riemann Equations as they seem to determine if a function is complex differentiable but I am not sure if my logic is sound. I created$$g\left(x,y\right)=a^{x+iy}$$$$=a^xa^{iy}$$$$=a^x\left(\cos \left(y\right)+\sin \left(y\right)i\right)$$$$=a^x\cos \left(y\right)+a^x\sin \left(y\right)i$$
From which I created $u\left(x,y\right)$ and $v\left(x,y\right)$
$$u\left(x,y\right)=a^x\cos \left(y\right)$$
$$v\left(x,y\right)=a^x\sin \left(y\right)$$
Which have partial derivatives
$$\frac{∂ u}{∂ x}=\ln \left(a\right)a^x\cos \left(y\right)$$
$$\frac{∂ v}{∂ y}=a^x\cos \left(y\right)$$
Which means that for $a\not=e\pm2π ni$
$$\frac{∂ u}{∂ x} \not= \frac{∂ v}{∂ y}$$
And so for $a\not=e±2π ni$ $f\left(x\right)$ is not complex differentiable. (Looking at $\frac{∂ v}{∂ x}$ and $\frac{∂ u}{∂ y}$ produces the same result so I have left that out here)
This seems reasonable but it is just not what I expected and I wanted to make sure that I haven't made any obvious mistakes. If you have any questions I will try to answer them to the best of my ability but as I said I have not actually been taught any of this so I might not know the answer.
 A: $f(z) = e^z$ is differentiable for all $z \in \Bbb C$. And if given any differentiable $f$ and constant $k, f'(kz) = kf'(z)$. Hence $e^{kz}$ is also differentiable.
The problem is not that $a^z$ is not differentiable for non-real or negative real $a$. The problem is that just saying "$a^z$" isn't sufficient to define a function.
We can define $e^z$ for complex $z$ by a number of different means. The most obvious is to define it by the Taylor series:
$$e^z := \sum_{n=0}^\infty \frac{z^n}{n!}$$
For other bases $a$, we can find a $k$ such that $a = e^k$, and define $a^z = e^{kz}$. The flaw in this is that $e^{k + 2n\pi i} = a$ as well for any integer $n$. So there is not one such value of $k$ but inifinitely many, and in general $e^{kz} \ne e^{(k + 2n\pi i)z}$, so you get different functions for different choices of the coefficient $k$.
Note that this holds even for positive real $a$. In fact, it holds even when $a = e$. For all of them, there are actually infinitely many exponential functions that we could define. However, for positive real $a$, exactly one of the potential values of $k$ is itself a real number, and we exploit this fact to select this real value of $k$ as the preferred one. Hence by convention (not because it is the only choice), we get $a^z = e^{z\ln a}$ for positive real $a$.
But when $a$ is negative or not real, there is no such preferred value of $k$. There remains an infinite number of possible definitions for $a^z$. Without additional information - something that specifies exactly which constant $k$ is being used - the function is not well-defined.
However, if you do finish the definition by choosing the coefficient $k$ (either directly or by imposing some condition that only one value of $k$ will satisfy), at that point $a^z = e^{kz}$ is well-defined and indeed is differentiable for all $z$.
