**UPDATE:** Previous argument was flawed. Here is what can salvage.

I can show there is no solution with $n$ an odd prime, or with $n$ odd and $\omega$ cyclotomic.

Let $\sigma$ denote complex conjugation.
Let $\zeta$ be a primitive $n$-th root of unity for $n$ odd and let $K = \mathbb{Q}(\zeta)$.

**Lemma:** Let $n$ be an odd prime and let $u$ be any unit of $K$, or let $n$ be odd and let $u$ be a cyclotomic unit of $K$. We have $\sigma(u)/u = \zeta^k$ for some integer $k$.

Proof: First, note that $\sigma(u)/u$ is an algebraic integer and all its Galois conjugates have norm $1$. So, by a result of Kronecker, it is a root of unity, and must be of the form $\pm \zeta^k$. Our goal is to show that the minus sign is impossible.

If $u$ is a cyclotomic unit, then it is a product of terms of the form $(1-\zeta^a)/(1-\zeta)$ and an explicit computation shows that the sign is positive.

Now, suppose that $n$ is an odd prime. So, suppose that $\sigma(u)/u = - \zeta^k$. Since $k$ is only determined modulo the odd number $n$, we may assume that $k$ is even. Replacing $u$ by $\zeta^{-k/2} u$, we have $\sigma(u)/u = -1$.

But, for any algebraic integer $v$ in $K$, we have $\sigma(v) \equiv v \mod 1-\zeta$. So $\sigma(u) \equiv u \mod 1- \zeta$ and (since $u$ is a unit) we have $\sigma(u)/u \equiv 1 \mod 1-\zeta$. Putting these together, we deduce that $1 \equiv -1 \mod 1-\zeta$. Since $n$ is an odd prime, $1-\zeta$ is a prime which does not divide $2$, a contradiction. $\square$

The error in the earlier version was forgetting that $1-\zeta$ can itself be a unit when $n$ is not prime. (In fact, this occur whenever $n$ is a square free non-prime.) And this unit, of course, violates the lemma. Not sure whether the original statement might still be true in these cases.

Now suppose that we have a solution to
$$\omega \equiv 1-\zeta^m \mod 4$$
for $m$ a proper divisor of $n$ and $\omega$ a unit. (I am using Franz's rephrasing.)

We hit both sides with $u \mapsto \sigma(u)/u$. By the lemma, we have $\sigma(\omega)/\omega = \zeta^j$ for some $j$. Also, $\sigma(1-\zeta^m)/(1-\zeta^m) = - \zeta^{-m}$.

So
$$\zeta^j \equiv - \zeta^k \mod 4$$
This equation is not true (using again that $n$ is odd), so we have a contradiction. $\square$

everychoice of $L$ or for just some choice of $L$? – KConrad Nov 25 '11 at 14:28onesuch field $L$ be enough for him or is he looking for infinitely many? Perhaps if he would explainwhyhe is asking this question it would be clearer what he really wants to know. – KConrad Nov 26 '11 at 2:20