Why functions on the complex plane are "analytically biased"?

Question

Why functions on the complex plane are "analytically biased"? Let me explain what Im thinking.

Observe that $$\frac{\mathrm d}{\mathrm dx}x^z=zx^{z-1},\quad z\in\Bbb C\setminus\{0\}$$

where $1\in\Bbb R$ instead of something like $$\frac{\mathrm d}{\mathrm dx}x^z=zx^{z-w},\quad z\in\Bbb C\setminus\{0\},w\in\Bbb S^1$$

The reason is that the binomial series is defined trough $k\in\Bbb N\implies k\in\Bbb R$, that is

$$\frac{\mathrm d}{\mathrm dx}x^z=\lim_{h\to 0}\frac{(x+h)^z-x^z}{h}=x^z\lim_{h\to 0}\left(-\frac1h+\sum_{k=0}^\infty\binom{z}{k}h^{k-1}t^{-k}\right)=x^z\cdot\frac{z}x$$

Then the functions $f:\Bbb C\to\Bbb C$ are analytically routed to the real line, so analytically $\Bbb C$ is not a symmetric plane. This is what I meant previously by "analytically biased".

So analysis seems not symmetric at all in a vector space despite of our intuitive notion of a plane to be "equal in each direction". This is right?

Or it is just our definition of derivative what is (naturally) routed to the real line?

There is some more deep reason for this asymmetry more than just history of mathematics and costume?

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EDIT:

By the comment of @user1952009 is clear that the derivative above can be seen as

$$\frac{\mathrm d}{\mathrm dx}x^z=\frac{\mathrm d}{\mathrm dx}\exp(z\operatorname{Log} x)=\frac{\mathrm d}{\mathrm dx}\exp(z(\ln |x|+i\operatorname{Arg}x))=x^zz(\frac1{|x|}+i\operatorname{Arg}'x)$$

where $\operatorname{Arg}(x):=\arg(x)+2\pi\Bbb Z$, so it seems that the derivative depends on the branch cut chosen for the complex logarithm. Then the origin of the asymmetry can be seen due to the branch cut?

Answer 1

Complex analytic functions are very rigid. One striking property they have is that if:

• $U$ is a connected, open domain
• $\{ x_n \} \to x$ is any sequence of points of $U$ converging to a point of $U$
• $f, g$ are analytic functions on $U$
• $f(x_n) = g(x_n)$ for all l$n$

then we can conclude $f = g$.

I'm pretty sure the same is true for multi-variable analytic functions.

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One application of this is that if the domain $U$ contains a real interval, then the value of any analytic function $f$ on $U$ is completely determined by its values on that real interval.

That is, if you have an analytic expression for the values of $f$ on the real interval and $U$ is contained in the domain of that expression, then the same analytic expression gives $f$ everywhere on $U$.

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By applying this to $f(x) = x^z$ (and taking care to deal with branch cuts), the identity $f'(x) = z x^{z-1}$ holds for positive real $x$ and real $z$, and thus it holds for all $x$ and $z$ where it makes sense.