 Solution

Choose the real embeddings ${\sigma }_1,\dots ,{\sigma }_r$ and one embedding ${\tau }_j$ from each pair of complex-conjugate embeddings. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition.

If $u$ lies in its kernel, every conjugate of $u$ has modulus one. Under the Minkowski embedding, every power $u^n$ therefore lies in one fixed bounded subset. Each $u^n$ belongs to ${\mathcal{O}}_K$, whose Minkowski image is discrete, so that bounded subset contains only finitely many such lattice points. Hence $u^a=u^b$ for some $0\le a<b$, andThus $u$ is a root of unity. Conversely, every root of unity clearly has all conjugates of modulus one and lies in the kernel. This proves the kernel of the logarithmic unit embedding is precisely $\mu (K)$.