$\begingroup$

The equation $N(z) = z \bar{z}$ with $N: \mathbf{Q}(i) \to \mathbf{Q}$ which implies $$\frac{1}{z} = \frac{\bar{z}}{N(z)}$$ generalizes to Galois extensions $L/K$: there is a field norm $$N_{L/K}(z) = \prod_{\sigma\in\operatorname{Gal}(L/K)} \sigma(z)\qquad \text{ with } \quad N_{L/K} : L \to K,$$ which implies $$\frac{1}{z} = \frac{w}{N_{L/K}(z)} \qquad \text{ where } \quad w = \prod_{\operatorname{id}

eq\sigma\in\operatorname{Gal}(L/K)} \sigma(z)\bigg.$$

In this case we have $L=\mathbf{Q}(\zeta_8)$, $K=\mathbf{Q}$, the minimum polynomial of $\zeta_8$ is $X^4 + 1$, and $$\operatorname{Gal}(L/K) = \{\zeta_8\mapsto\zeta_8,\ \ \zeta_8\mapsto\zeta_8^3,\ \ \zeta_8\mapsto\zeta_8^5 = -\zeta_8,\ \ \zeta_8\mapsto\zeta_8^{7} = -\zeta_8^3\}.$$

So for $z= \alpha+\beta\zeta_8+\gamma\zeta_8^2+\delta\zeta_8^3$ you have the pleasure of doing this calculation.

To save time we can do the calculation in SageMath:

L.<zeta8> = CyclotomicField(8) R.<a,b,c,d> = PolynomialRing(L) z = a + b*zeta8 + c*zeta8^2 + d*zeta8^3 w = prod(z.map_coefficients(sigma) for sigma in L.galois_group() if not sigma.order() == 1) N = w*z w_vect = sum(vector(R,p)*q for (p,q) in list(w))

From the values of w_vect and N we see $$w = (a^3 - b^2c + ac^2 + 2abd + cd^2) + (-a^2b + bc^2 - b^2d - 2acd - d^3)\zeta_8 + (ab^2 - a^2c - c^3 + 2bcd - ad^2)\zeta_8^2 + (-b^3 + 2abc - a^2d + c^2d - bd^2)\zeta_8^3$$ and $N_{\mathbf{Q}(\zeta_8)/\mathbf{Q}}(z) = a^4 + b^4 - 4ab^2c + 2a^2c^2 + c^4 + 4a^2bd - 4bc^2d + 2b^2d^2 + 4acd^2 + d^4.$

Let's check to be sure:

sage: (w_vect[0] + w_vect[1]*zeta8 + w_vect[2]*zeta8^2 + w_vect[3]*zeta8^3)*z == N True

An example, as requested by Mr. Brooks: to calculate $\frac{\zeta_8-\zeta_8^3}{\zeta_8+\zeta_8^3}$ we put $z = \zeta_8+\zeta_8^3$ and compute $$\begin{align*}w &= (\zeta_8^3+\zeta_8^9)(-\zeta_8-\zeta_8^3)(-\zeta_8^3-\zeta_8^9)\\ &= (\zeta_8+\zeta_8^3)^3\\&=\zeta_8^3 + 3\zeta_8^2\zeta_8^3 + 3\zeta_8\zeta_8^6 + \zeta_8^9\\&= \zeta_8^3 - 3\zeta_8 - 3\zeta_8^3 + \zeta_8\\ &= -2(\zeta_8 + \zeta_8^3)\end{align*}$$ and $$N_{L/K}(z) = wz = -2(\zeta_8 + \zeta_8^3)^2 = -2(\zeta_8^2 + 2\zeta_8\zeta_8^3 + \zeta_8^6) = -2(\zeta_8^2 - 2 - \zeta_8^2) = 4$$ so $$\frac{1}{z} = \frac{w}{N_{L/K}(z)} = -\frac{1}{2}(\zeta_8+\zeta_8^3).$$ (This is the same as putting $(a,b,c,d) = (0,1,0,1)$ in the formulae above.)

Finally: $$\frac{\zeta_8-\zeta_8^3}{\zeta_8+\zeta_8^3} = (\zeta_8-\zeta_8^3)\cdot\frac{1}{z} = -\frac{1}{2}(\zeta_8-\zeta_8^3)(\zeta_8+\zeta_8^3) = -\frac{1}{2}(\zeta_8^2 - \zeta_8^6) = -\zeta_8^2.$$

All the above holds in an abstract number field $L= \mathbf{Q}(\zeta_8)$ where $\zeta_8$ is a root of $X^4+1$. If you choose the embedding $L\to\mathbb{C}$ given by $\zeta_8 \mapsto \exp(2\pi i/8)$ then e.g. the example computes $$\frac{\sqrt{2}}{\sqrt{-2}} = -i.$$