= Solution
The multiplicative <Hilbert theorem 90> states that, for a finite <Galois extension> $L/K$ with <Galois group> $G$, every multiplicative <cocycle> $c:G\to L^\times$ satisfying $c_{\sigma\tau}=c_\sigma\sigma(c_\tau)$ has the form $c_\sigma=\sigma(y)/y$ for some $y\in L^\times$. In the cyclic case $G=\langle\sigma\rangle$, it says
$$
\boxed{N_{L/K}(x)=1\iff x=\sigma(y)/y\text{ for some }y\in L^\times.}
$$
Here is the full existence argument. Distinct <field homomorphisms> $\sigma:L\to L$ are linearly independent as functions over $L$. To see this <Artin independence theorem> directly, assume a nonzero identically vanishing relation with the fewest nonzero coefficients. There are at least two terms. Choose two participating embeddings $\sigma_1,\sigma_j$ and an element $a$ with $\sigma_1(a)\ne\sigma_j(a)$. Evaluate the relation at $at$ and subtract $\sigma_j(a)$ times its value at $t$. The $j$th term disappears while the first remains nonzero, giving a shorter vanishing relation, a contradiction.
Apply this independence to
$$
A(t)=\sum_{\sigma\in G}c_\sigma^{-1}\sigma(t).
$$
The coefficients are nonzero, so $A$ is not identically zero; choose $t$ with $y=A(t)\ne0$. The <cocycle> identity gives $\tau(c_\sigma^{-1})=c_{\tau\sigma}^{-1}c_\tau$, and hence
$$
\tau(y)=\sum_{\sigma\in G}\tau(c_\sigma^{-1})\tau\sigma(t)
=c_\tau\sum_{\rho\in G}c_\rho^{-1}\rho(t)=c_\tau y.
$$
This proves the general statement. In the cyclic case, given $N(x)=1$, set $c_1=1$ and $c_{\sigma^j}=x\sigma(x)\cdots\sigma^{j-1}(x)$ for $1\leq j<n$. The norm-one identity makes this definition compatible with $\sigma^n=1$, and the <cocycle> identity follows by concatenating the products. The argument above gives $\sigma(y)/y=x$. Conversely that quotient has norm one because its conjugate product telescopes.
Finally let $L/K$ be <unramified>, and take its <arithmetic Frobenius> $\phi$. For a unit $x$ of norm one, the cyclic theorem gives $b\in L^\times$ with $\phi(b)/b=x$. Write $r=v_L(b)$ and choose the common <uniformizer> $\pi\in K$. Then $y=\pi^{-r}b$ has valuation zero, and $\phi(\pi)=\pi$ implies $\phi(y)/y=\phi(b)/b=x$. Thus the <unit form of Hilbert theorem 90> is
$$
\boxed{x\in\mathcal O_L^\times,\ N_{L/K}(x)=1\Longrightarrow\exists y\in\mathcal O_L^\times:\ \phi(y)/y=x.}
$$
The unramified hypothesis is what permits shifting an arbitrary integer valuation of $b$ by a power of a uniformizer belonging to $K$.
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