First of all, I realized that I had made a big mistake. The Galois extension $k_{n}/k^{+}$ is abelian (since it is a composite of $k/k^{+}$ and $\mathbb{B}_{m}/\mathbb{Q}$ for some m.) hence the complex conjugation commutes with any element of $G(k_{n}/k)$.
And the first question on existence of roots of unity $w$ $\in k_{n}$ such that $N_{n,0}(w) = N_{n,0}(\beta)$ was right as the author claimed and was actually important.
In general, for a number field $k$ and the nth-layer field $k_{n}$ of the cyclotomic $\mathbb{Z}_{l}$-extension, for any root of unity $\alpha \in k$, there exists a root of unity $w \in k_{n}$ such that $N_{n,0}(w)=\alpha$.
If the multipative order of $\alpha$ is prime to $l$, then as $\alpha$ is invariant under Galois action, the existence of $w$ is trivial. Hence we can assume that the multiplicative order of $w$ is a power of $l$.
If the order is strictly larger than 1, then $k$ contains $\mathbb{Q}(\mu _{l})$. By the theory of $\mathbb{Z}_{l}$-extension, the intersection of $k$ and "the cylclotomic $\mathbb{Z}_{l}$ extension of $\mathbb{Q}$" is $\mathbb{Q}(\mu _{l^p})$ for some $p$, which means $k_{n}$ is the extension field of $k$ generated by adjoining $l^{p+n}$th roots of unity.
Then we're done.