= 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 $e^{i\varepsilon}$ for small positive $\varepsilon$ has the two preimages with arguments $\varepsilon/4$ and $\pi/2+\varepsilon/4$. 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 $\Delta$.
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