Class number and Narrow Class Number It is well documented that there are nine imaginary quadratic number fields with class number 1 and hence they have narrow class number 1 as well. However, are there imaginary quadratic number fields with class number > 1 and narrow class number 1? If so, are there lots of them?
 A: The narrow class number of a number field $K$ is just the cardinality of the corresponding narrow class group $\operatorname{Cl^+}(K) = I(K)/P^+(K)$ where $P^+(K)$ is the group of principal fractional ideals $(\alpha) = \alpha \mathcal{O}_K$ whose generator $\alpha \in K$ is totally positive.
Now, this is basically the important part here, because $\alpha$ is totally positive if for every embedding $\sigma: K \hookrightarrow \mathbb{R}$, the image of $\alpha$ under the embedding $\sigma(\alpha) > 0$.
Then in the case you're interested in, say when $K = \mathbb{Q}(\sqrt{d})$ is an imaginary quadratic field, there are no embeddings $\sigma: K \hookrightarrow \mathbb{R}$, thus the condition for a principal fractional ideal $(\alpha)$ to belong to $P^+(K)$ is trivially satisfied, therefore $P^+(K) = P(K)$, the whole group of principal fractional ideals. 
Thus in the case when $K$ is an imaginary quadratic number field, or more generally a number field with no real embeddings, the narrow class group and the class group are actually the same and thus the class number and the narrow class number are also the same.
Added: Matt E's comment
As Matt E comments, the ideal class group $\operatorname{Cl}(K)$ is a quotient of the narrow class group by virtue of the third isomorphism theorem for groups, that is
$$
\operatorname{Cl}(K) = I(K)/P(K) \cong \frac{\frac{I(K)}{P^+(K)}}{\frac{P(K)}{P^+(K)}} = \frac{\operatorname{Cl^+}(K)}{\frac{P(K)}{P^+(K)}}
$$
and therefore as he observes in his comment below, if the narrow class group $\operatorname{Cl^+}(K)$ has order $1$ then as a consequence the class group $\operatorname{Cl}(K)$ also has order $1$ and thus 
$$\text{narrow class number} = 1 \implies \text{class number} = 1$$
