Why $f(z)=z^2$ is single valued? Why $f(z)=z^2$ is single valued where $z\in\mathbb{C}$? From definition we have $$z^2=e^{2 \log z}=e^{2(\ln|z|+i(2k\pi+Arg(z)))}$$ I dont get it ;/ Maybe it's getting late.

It is a matter of definition.

Normally, $z^a$ with $a \in \mathbb Z$ _can_ be defined as a single-valued function without drawbacks, so we do so; it is not defined as $\exp(a \log z)$ but as $z \cdot \dots \cdot z$ $a$ times. (or if $a$ is negative $\frac 1z \cdot \dots \cdot \frac 1z$ $a$ times)

Instead, $z^a$ with $a \in \mathbb R - \mathbb Z$ admits multiple values; in particular, if $a = m/n$ it has $n$ distinct values, while if $a$ is irrational then it has infinite multiple values. (This comes of course from the fact that the complex logarithm is a polydrome function)

There is no ambiguity though, because if $a \in \mathbb Z$ we have that $\exp(a \log z) = z^a$ (as defined above); in particular $\exp(a \log z)$ admits a single value.