Dirichlet's unit theorem gives all units of an algebraic number field $\mathbb{Q}(\vartheta)$ in the unique form $$\varepsilon = \zeta^{n}\eta_1^{k_1}\eta_2^{k_2}...\eta_t^{k_t},$$ where $\zeta$ is a primitive $w^\mathrm{th}$ root of unity in $\mathbb{Q}(\vartheta)$ the $\eta_j$ s are the fundamental units of $\mathbb{Q}(\vartheta)$ $0 \leqq n \leqq w\!-\!1$ $k_j \in \mathbb{Z}$ , $\forall j$ $t = r\!+\!s\!-\!1$

• The case of a real quadratic field $\mathbb{Q}(\sqrt{m})$ the square-free $m > 1$ $r = 2$ $s = 0$ $t = r\!+\!s\!-\!1 = 1$ So we obtain $$\varepsilon = \zeta^{n}\eta^{k} = \pm\eta^{k},$$ because $\zeta= -1$ , is the only real primitive root of unity ($w = 2$ . Thus, every real quadratic field has infinitely many units and a unique fundamental unit $\eta$

  Examples: If $m = 3$ then $\eta = 2\!+\!\sqrt{3}$ if $m = 421$ then $\eta = \frac{444939+21685\sqrt{421}}{2}$

• The case of any imaginary quadratic field $\mathbb{Q}(\vartheta)$ here $\vartheta = \sqrt{m}$ the square-free $m < 0$ The conjugates of $\vartheta$ are the pure imaginary numbers $\pm\sqrt{m}$ hence $r = 0$ $2s = 2$ $t = r\!+\!s\!-\!1 = 0$ Thus we see that all units are $$\varepsilon = \zeta^{n}.$$

  1) $m = -1$ The field contains the primitive fourth root of unity, e.g. $i$ and therefore all units in the Gaussian field $\mathbb{Q}(i)$ are $i^n$ where $n = 0,\,1,\,2,\,3$

  2) $m = -3$ The field in question is a cyclotomic field containing the primitive third root of unity and also the primitive sixth root of unity, namely $$\zeta = \cos{\frac{2\pi}{6}}+i\sin{\frac{2\pi}{6}};$$ hence all units are $\varepsilon = (\frac{1+\sqrt{-3}}{2})^{n}$ where $n = 0,\,1,\,\ldots,\,5$ or, equivalently, $\varepsilon = \pm(\frac{-1+\sqrt{-3}}{2})^{n}$ where $n = 0,\,1,\,2$

  3) $m = -2$ $m <-3$ The only roots of unity in the field are $\pm 1$ hence $\zeta = -1$ $w = 2$ and the units of the field are simply $(-1)^{n}$ where $n = 0,\,1$