 Solution

Here “covering map” uses the course convention of a local homeomorphism; “regular” adds the evenly-covered product condition. The derivative of $p(z)=z^4$ is $4z^3$, which never vanishes on the upper half-plane $\mathbb{H}$. The holomorphic inverse function theorem therefore makes $p$ a local biholomorphism and hence a covering map in this sense.

It is not regular. The point $1$ has only the preimage $i$ in $\mathbb{H}$, whereas $e^{i\epsilon }$ for small positive $\epsilon$ has the two preimages with arguments $\epsilon /4$ and $\pi /2+\epsilon /4$. Thus the fibre cardinality is not locally constant near $1$, so no neighbourhood of $1$ has a preimage of the form $U\times \Delta$ for a fixed discrete set $\Delta$.