Is there a purely imaginary unit in the cyclotomic number field of an odd prime degree? Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th  root of unity in $\mathbb{C}$.
Let $K = \mathbb{Q}(\zeta)$.
Let $A$ be the ring of algebraic integers in $K$.
My question: Is there a purely imaginary unit in $A$?
EDIT
New question: Is the following proposition true? If yes, how would you prove this?
Proposition
There is no purely imaginary unit in $A$.
Related questions:
On a certain property of the different of an extension of an algebraic number field of a prime relative degree
Maximal real subfield of Q(ζ)
 A: Not necessarily: if $\ell = 3$ then there are only six units in $A$ and none of them are totally imaginary.
A: The proposition is true:
Let $K^+$ denote the totally real subfield of $K$.    Let $U$ be the group of units in $K$, and $U^+$ be the group of units in $K^+$.  Dirichlet's Theorem
implies that $U^+$ has finite index in $U$.  I claim that in fact the index
is odd.  Indeed, suppose that $u$ is an element of whose image in $U/U^+$ is of exact order $2$.
Then $K = K^+(u) = K^+(\sqrt{u^2})$ is obtained by extracting the square root of a unit, and so is unramified over $K^+$ except possibly at primes lying over $2$.  However, we know that $K/K^+$ is ramified at precisely the prime lying over $l$.
Consequently $U/U^+$ has odd order.  
If $u \in U$ were purely imaginary, then $u \not\in U^+$,
but $u^2 = - u \overline{u} = - |u|^2$ is an element of $U^+$, contradicting what we have just proved.  Thus $U$ contains no purely imaginary elements.
[Hopefully this is correct; the previous argument I posted was nonsense.]
