= Solution
First suppose $|\sigma_i(\alpha)|=1$ for every complex embedding. The coefficients of the monic polynomial
$$
P_m(T)=\prod_{i=1}^n(T-\sigma_i(\alpha^m))
$$
are rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers $\alpha^m$ occur. Two powers coincide; since $\alpha\neq0$, this gives $\alpha^{r-s}=1$. Thus $\alpha$ is a <root of unity>. This is <Kronecker theorem on algebraic integers in the unit disk>.
The second printed assertion is literally false for $\alpha=0$. For the intended statement with $\alpha\neq0$, use the <Minkowski embedding of a number field>[Minkowski embedding]: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one has
$$
1\leq|N_{K/\mathbb Q}(\alpha)|leq1,
$$
so every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has $\max_i|\sigma_i(\alpha)|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha\in\mathcal O_K$ satisfying $|\sigma_i(\alpha)|<c$ for all $i$ is a root of unity.
<Dirichlet unit theorem> states that for signature $(r_1,r_2)$,
$$
\mathcal O_K^\times\cong\mu(K)\times\mathbb Z^{r_1+r_2-1},
$$
and the <logarithmic embedding of number field units> maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.
For a real quadratic field, the roots of unity are $\{\pm1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb Z\subset\mathbb R$. Its smallest positive element is $\log\epsilon$ for a unit $\epsilon>1$. Given any positive unit $u>1$, choose $n$ so that $1\leq u\epsilon^{-n}<\epsilon$; minimality forces the quotient to be $1$. Applying signs and inverses shows
$$
\mathcal O_K^\times=\{\pm\epsilon^n:n\in\mathbb Z\}.
$$
Since $11\equiv3\pmod4$, the <ring of integers of a quadratic field> is $\mathbb Z[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the <Pell equation> $a^2-11b^2=\pm1$. The element
$$
10+3\sqrt{11}
$$
has norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\leq3$. For $b=1,2,3$, the numbers $11b^2\pm1$ are respectively
$$
10,12;qquad43,45;qquad98,100.
$$
Only the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and
$$
\boxed{\epsilon=10+3\sqrt{11}}.
$$
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