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Let $H_k$ be the ordinary Hilbert class field of $k$. Its real places split completely, so $H_k$ is totally real. Since $K/k$ is quadratic and $K$ is totally imaginary, $H_k\cap K$ cannot equal $K$; hence it equals $k$. Therefore the compositum $K{H}_k$ hasand is an abelian extension of $K$, by restriction of the Galois group to $H_k$. At finite primes it is unramified because unramified extensions remain unramified under base change. The field $K$ has no real places, so there is no infinite-place ramification condition left. Thus $K{H}_k\subseteq H_K$, where $H_K$ is the Hilbert class field of $K$. Since $[H_K:K]=h_K$, the tower degree formula proves class number divisibility for a CM extension:This argument uses the ordinary class field with split real places; replacing it silently by the narrow class field would invalidate the claim that $H_k$ is totally real.