= Solution
The anti-muon has the same charge as a proton, so the Coulomb potential seen by the electron is the same as in the <hydrogen atom>. The main difference is the source mass. Separating the centre-of-mass motion replaces $m_e$ by the <Coulomb two-body reduced mass>
$$
\mu=\frac{m_em_\mu}{m_e+m_\mu}.
$$
Since $m_\mu\gg m_e$, this is close to $m_e$, though less close than the electron-proton reduced mass. The smaller reduced mass gives a slightly larger Bohr radius and energies of slightly smaller magnitude. Under the assumptions in the question, the Coulomb form, angular eigenfunctions and quantum numbers are otherwise unchanged; anti-muon decay and spin effects are being neglected.
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