Norm of $2^{i}$-th primitive root

Question

Let $ K $ be finite degree extension of $ \mathbb{Q} $ such that $ -1 $ is not a square in $ K $. Let $ L = \frac{K[x]}{\langle x^2 +1\rangle}$. Thus every element of $ L $ is of the form $ a + ib $ where $ i^2 = -1 $ and $ a,b \in K $. Let $ \eta_{2^i} $ is the primitive $ 2^i $-th root of unity contained in $ L $ where $ i \geq 2 $ . Can $ N_{L/K}(\eta_{2^i})$ be $ -1 $? Where $ N_{L/K} $ is a norm map from $L$ to $K$ and defined by $ N_{L/K}(b) = b \sigma(b) $ for all $ b \in L $ where $ \sigma $ is the automorphism of the $ \operatorname{Gal}(L/K)$. If $ L $ is real subfield then it is clear.

Answer 1

Let $K=\mathbb{Q}[\sqrt{-2}]$. Then $\frac{1}{2}\sqrt{-2}(1-i)$ is an eighth root of unity in $L$ whose norm is $\frac{1}{2}\sqrt{-2}(1-i)\frac{1}{2}\sqrt{-2}(1+i)=\frac{1}{4}(-2)(1-i^2)=-1\in K$. [I'm assuming that when you write $2^i$ it's a different $i$]