Why is a function like $z^{2.5}$ not holomorph?

Question

If I consider the function $f(z)=z^{2.5}$ $\in \mathbb{R}$, it is continuously differentiable with $f'(z)=2.5z^{1.5}$. But why can I not do the same in $\mathbb{C}$ with $ f'(z)=2.5z^{1.5}$?

Answer 1

The function $f(z)=z^{2.5}$ is only well defined on positive real numbers. Everywhere else, you need to provide additional info. For example, it is not clear what $(-1)^{2.5}$ would be equal to.

You could write the function as $z^{2.5} = e^{2.5\cdot \log(z)}$, but then, the function $\log$ is not well defined on the complex numbers. Taking the standard logarithm, the function $\log$ is undefined in $(-\infty, 0]$, and there is no way you can ever continuously define the logarithm function on the entire complex plane.