I'm a little insecure about determination of logarithms and square roots and I don't know if it's alright:
Let $f(z)=z(z-1)$, $z\in \mathbb{C} \setminus [0,1]$.
I have to prove that there is no determination of logarithm but there is a determination of the square root.
If $Arg(z)$ is the principle argument (that goes from $-\pi$ to $\pi$) it is known that it doesn't have a continuous determination in $\mathbb{C}\setminus {\{0\}}$ so when the argument of $z$ makes a round to the complex plane then the argument of $f(z)$ is supposed to make more than a round. As we can see there is not a continuous determination of $f(z)$ and as long as the logarithm is expressed as: $$Log(z)=|z|+i Arg(z) $$ we do not have a logarithm determination.
Nevertheless when we are talking about the root square the argument isn't doing a full complete round to the complex plane so we don't have this continuity problem because there isn't any number in $\mathbb{C}\setminus (-\infty,0]$, so there is a continuous determination of the root square. Thank you very much.
If $f$ had a logarithm $l$, then$$l'(z)=\frac{f'(z)}{f(z)}=\frac{1-2z}{z-z^2}.$$But$$\operatorname{res}_{z=0}\left(\frac{1-2z}{z-z^2}\right)=\operatorname{res}_{z=1}\left(\frac{1-2z}{z-z^2}\right)=1$$and therefore$$\oint_{\lvert z\rvert=2}\frac{1-2z}{z-z^2}\,\mathrm dz=4\pi i.\tag1$$So, there is no such $l$; if there was, since it would be a primitive of $\frac{1-2z}{z-z^2}$, the integral $(1)$ should be equal to $0$.