Is there any approximated version of Hilbert 90? Suppose $K$ is a local field and $L$ a finite cyclic extension of $K$. By Hilbert 90, we know that if an element $a$ in $ L$ such that $N_{L / K}(a) =1 $ then $ a = b / \sigma(b) $ for some $b$ in $L$ and $ \sigma$ a generator of the Galois group.
My question: Suppose $N_{L / K}(a) \simeq 1 $, is there some $b$ such that  $ a \simeq b /\sigma(b) $?
I guess I could have try to make the question more precise, but there seems to be some merits in leaving it a bit vague.
I have accept the the answer by Paul Broussous, which address the situation when the extension is unramified. It is because that is what I need. I am still curious whether something can be done when the extension is totally ramified?
 A: Yes such approximated versions of Hilbert 90 do exist. But you need some technical conditions.
For instance assume that $L/K$ is unramified of degree $d$ and that 
$a\in {\mathfrak o}_{L}^{\times}$. 
Then you condition writes 
$N_{L/K}(a)\equiv 1$  modulo  $\mathfrak{p}_{K}^{n}$,
for some $n>0$ (I assume that this 
is what you mean by $\simeq$). This may be rewritten 
$N_{L/K}(a)=1$ in $U_{L}/U^{n}_L$, where $U$ denotes a 
unit group. So the map 
$$\sigma^u\mapsto a\sigma (a) \cdots \sigma^u (a)$$
 defines a $1$-cocycle of ${\rm Gal}(L/F)$ in $U_L /U^n_L$  (here $\sigma$ denotes the Frobenius substitution). 
So what you want is that this cocycle is split. In fact we have $H^{1}({\rm Gal}(L/K), U_{L}/U^{n}_{L})=1$. This is proved by a standard filtration argument: this is implied by 
$$H^{1}({\rm Gal}(L/K), U_L /U^{1}_L ) = H^{1}({\rm Gal}(k_L /k_K ), k_{L}^{\times})=1$$
and 
$$H^{1}({\rm Gal}(L/K), U_{L}^{i}/U_{L}^{i+1})=H^{1}({\rm Gal}(k_L /k_K ), k_L )=1$$
here $k$ denotes a residue field. You can find the detail of the proof in, I think,  Serre's 'Local fields' or Cassels-Fröhlich's 'Algebréaic Number Theory'. 
