The log part is answered in the question Why is there no continuous log function on $\mathbb{C}\setminus\{0\}$?, so I will address only the square root.
Suppose, to reach a contradiction, that $f:\mathbb C\setminus\{0\}\to\mathbb C$ is a continuous square root function, i.e. $f(z)^2 = z$ for all $z\neq 0$. The restriction of $f$ to the unit circle $\mathbb T$ maps $\mathbb T$ to itself, and it is injective and continuous. Because $\mathbb T$ is compact and Hausdorff, this implies that $f(\mathbb T)$ is homeomorphic to $\mathbb T$. Since $f(\mathbb T)$ is compact and connected the only possibilities for $f(\mathbb T)$ are a proper closed arc or all of $\mathbb T$, and the former is ruled out because it is not homeomorphic to $\mathbb T$ (e.g., because from a proper closed arc you can remove 2 points without losing connectedness).
Hence $f(\mathbb T)=\mathbb T$, so there exist $w$ and $z$ such that $f(w)=-1$ and $f(z)=1$. This implies that $w=f(w)^2=(-1)^2=1$, and $z=f(z)^2=1^2=1$. So $z=w$, but $f(z)\neq f(w)$, which is absurd. Thus such an $f$ cannot exist.
(If instead you wanted to restrict to a small punctured disk, the same argument applies to $f$ mapping $|z|=r$ to $|z|=r^2$.)