Roots of unity of a quadratic field Let $K$ be a quadratic number field, and $\mu(K)$ the group of roots of unity of $K$. I'm trying to prove that (with the two exceptions of $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-3})$)
$$\mu(K) = \{1, -1\}$$
The case where $K$ is imaginary is easier, since by Dirichlet's unit theorem, we have
$$\mathcal{O}_K^* = \mu(K) \times \mathbb{Z}^{1 + 0 - 1} = \mu(K)$$
And it is easy to prove that $\alpha \in \mathcal{O}_K^* \iff N(\alpha) = 1$. Dealing with the $d \equiv 2, 3 \mod 4$ and $d \equiv 1 \mod 4$ separately we get a complete characterization of $\mathcal{O}_K^*$, and therefore of $\mu(K)$, for $K$ imaginary.
But for $K$ real this doesn't work anymore, because we have
$$\mathcal{O}_K^* = \mu(K) \times \mathbb{Z}^{0 + 2 - 1} = \mu(K) \times\mathbb{Z}$$
Instead, and the approach above gives infinite units (solution to Pell's equation), and says nothing about the roots of unity.

I'm guessing that one could use the fact that a real quadratic field has only one embedding, and that embedding is real, together with the fact that the only roots of unity in $\mathbb{R}$ are $\pm 1$.

Is there an elementary way to prove the result without separating the
  real and imaginary case, and/or without using Dirichlet's unit theorem?

 A: The real case is actually quite easy! Consider that the torsion subgroup is $\{\alpha : \alpha^n =1\}$ since the group is multiplicative. But then this is exactly the roots of unity. As such you need only note that since $F\subseteq\Bbb R$ we have that only real roots of unity are in this field, i.e. $\{\pm 1\}$.
If you care about the general case all together, it's also easy, again you note that $\mu(K)$ is exactly the roots of unity since torsion iff finite multiplicative order iff root of unity. But then if you have a root of unity, $\zeta_n$, for some $n\ne 2,3,4,6$ (recall $-\zeta_3$ is a primitive $6^{th}$ root of unity) then you have a proper subfield of degree $\phi(n)$. Since $p-1|\phi(n)$ for all $p|n$ we see $n=2^\alpha3^\beta$ otherwise we have a sub-extension of too large a degree. But we have excluded the (non-trivial) cases

$$\begin{cases}
\alpha = \beta = 1\\
 \alpha=2, \beta = 0 \\
\alpha = 0, \beta = 1
\end{cases}$$

And if $\alpha >2 $ or $\beta >1$ the degree is again too big since then $\phi(n)>2$, so no other cases can occur.
A: If $p$ is a prime then $\zeta_p$ satisfies an irreducible equation of degree $p-1$. Since quadratic fields have degree $2$, it remains only to consider the cube roots and the power of $2$ roots. 
