Exercise 3.1.9 (Homorphism $\varphi : \mathbb{C}^\times \to \mathbb{R}^\times$ defined by $\varphi(a +bi) = a^2 + b^2$)

Proof.

Note first that if $a+\mathit{bi}\in {ℂ}^{\text{×}}$, then $(a,b)\ne (0,0)$, thus $a^2+b^2\in {ℝ}^{\text{×}}$ so $\phi :{ℂ}^{\text{×}}\to {ℝ}^{\text{×}}$ is well defined.

If $z=a+\mathit{bi}\in ℂ$, where $a,b\in ℝ$, then

Note that if $z=a+\mathit{bi},t=c+\mathit{di}$, ( $a,b,c,d\in ℝ$) are complex numbers, then

(So the map defined by $z\mapsto \bar{z}$ is an automorphism of the group ${ℂ}^{\text{×}}$.)

Therefore, for all $z,t\in {ℂ}^{\text{×}}$

so $\phi :{ℂ}^{\text{×}}\to {ℝ}^{\text{×}}$ is a homomorphism.

If $c\in \mathrm{im}(\phi )$, then $c=\phi (a+\mathit{bi})=a^2+b^2>0$ for some reals $a,b$, so $c\in {ℝ}_{\text{+}}^{\text{×}}$.

Conversely, if $c\in {ℝ}_{\text{+}}^{\text{×}}$, then $c={(\sqrt{c})}^2+0^2=\phi (\sqrt{c}+0\cdot i)$, where $(\sqrt{c},0)\in {ℝ}^2$, so $c\in \mathrm{im}(\phi )$. So

For every $a\in \mathrm{im}(\phi )={ℝ}_{\text{+}}^{\text{×}}$, the fiber above $a$ is

corresponds in the Euclidean plane ${ℝ}^2$ to the circle with center $(0,0)$ and radius $\sqrt{a}$.

In particular, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ is the subgroup $𝕌$ corresponding in the plane to the circle with center $(0,0)$ and radius $1$. □