Well, actually the kernel of $\theta$ is perfectly explicit and it is the connected component of the identity in $C_K$: see, for instance, Artin-Tate Class Field Theory, Chapter IX, §1. Theorem 3 ibid says
Theorem 3 The structure of the connected component of $C_K$ is that of a direct product of one real line $\mathbb{R}_+$, $r=r_1+r_2-1$ solenoids and $r_2$ circles $\mathbb{T}$.
As usual, $r_1$ is the number of real embeddings, $r_2$ that of conjugate pairs of complex ones, and the solenoid is the quotient $\widehat{\mathbb{Z}}\times\mathbb{R}/\mathbb{Z}$$\mathbb{S}=(\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Z}$: this is homeomorphic to the dual group of $\mathbb{Q}$, when the rationals have the discrete topology.
When $K=\mathbb{Q}$, so that $r_2=r=0$, you find the result that you quote in your question. In general, you see that the answer is independent of the discriminant (or of any other deep arithmetic property of the field, for that matter): only the number of real/complex places plays a role. In particular, the answer to your question is: "the kernel is $\mathbb{R}_+$ only if $K=\mathbb{Q}$", because for every other number field $K$, $r>0$ unless $K$ is imaginary quadratic, in which case $r_2=1$.
Added The above discussion implies that the connected component of $C_K$ (i. e. the kernel of $\theta$) is isomorphic to $\mathbb{R}_+^i\times(\mathbb{C}^\times)^j$ only when $K=\mathbb{Q}$ or when $K$ is imaginary quadratic: this is equivalent to saying that $r=0$. To prove the claim, it is enough to show that $$ \mathbb{R}_+^i\times(\mathbb{C}^\times)^j\cong \mathbb{R}_+\times \mathbb{S}^r\times \mathbb{T}^{r_2}\Longrightarrow r=0. $$ Now observe that both topological groups $\mathbb{R}_+^i\times(\mathbb{C}^\times)^j$ and $\mathbb{R}_+\times \mathbb{S}^r\times \mathbb{T}^{r_2}$ are locally compact (the fact that $\mathbb{S}$ is compact is in Artin-Tate, Chapter IX, §1, Lemma 2) and connected. It follows (see, for instance, this page in the nLab) that both admit a maximal compact subgroup, and an isomorphism as above would send the maximal compact subgroup of one, to that of the other. Moreover, to check whether a compact subgroup is maximal, it is enough to check whether the quotient by it is contractible. It follows that the maximal compact subgroup of $\mathbb{R}_+^i\times(\mathbb{C}^\times)^j \approx\mathbb{R}_+^{i+j}\times\mathbb{T}^j$ is homeomorphic to $\mathbb{T}^j$, while the maximal compact subgroup of $\mathbb{R}_+\times \mathbb{S}^r\times \mathbb{T}^{r_2}$ is homeomorphic to $\mathbb{S}^r\times\mathbb{T}^{r_2}$. We now need to show that this forces $r=0$: this follows from the structure theorem of divisible groups. Indeed, the same Lemma 2 quoted above says that $\mathbb{S}$ is uniquely divisible, hence a $\mathbb{Q}$-vector space; on the other hand $\mathbb{T}$ is certainly divisible, but not uniquely. In particular, a group of the form $\mathbb{T}^j$ cannot contain any infinitely divisible subgroup, by the structure theorem, and cannot be isomorphic to $\mathbb{S}^r\times\mathbb{T}^{r_2}$ unless $r=0$. This finishes the argument.
Finally, observe that when $K=\mathbb{Q}$, then $i=1$ and $j=0$ (as you have already noticed); while when $K$ is imaginary quadratic, the above computation shows $i=0$ and $j=1$ (this ultimately requires that $\mathbb{R}_+^a\cong \mathbb{R}_+^b\Rightarrow a=b$).