Question # Show that displaystyle{C}^{ast}={R}^{ast}+times{T},text{where} {C}^{ast} is the

Vectors and spaces
ANSWERED Show that $$\displaystyle{C}^{\ast}={R}^{\ast}_{+}\times{T},\text{where}\ {C}^{\ast}$$ is the multiplicative group of non-zero complex numbers, T is the group of complex numbers of modulus equal to 1, $$\displaystyle{R}^{\ast}_{+}$$ is the multiplicative group of positive real numbers. 2021-03-05

Recall that each nonzero complex number z can be written (uniquely!) in the form
$$\displaystyle{z}={\left|{z}\right|}{e}^{{{i}\alpha}}.$$
where $$\displaystyle{\left|{z}\right|}$$ is the modulus of z, and $$\displaystyle\alpha\in{\left[{0},{2}\pi\right\rangle}$$ is the argument of z. Also,
$$\displaystyle{e}^{{{i}\alpha}}$$ is an element of TT, and $$\displaystyle{\left|{z}\right|}$$ is a real positive number (positive since $$(\displaystyle{z}\ne{0})$$)
Furthermore, $$\displaystyle\le{t}{z}_{{1}}{\quad\text{and}\quad}{z}_{{2}}$$ be two nonzero complex numbers and
$$\displaystyle{z}_{{1}}={\left|{z}_{{1}}\right|}{e}^{{{i}\alpha_{{1}}}}$$
$$\displaystyle{z}_{{2}}={\left|{z}_{{2}}\right|}{e}^{{{i}\alpha_{{2}}}}$$
Then
$$\displaystyle{z}_{{1}}{z}_{{2}}={\left|{z}_{{1}}\right|}{e}^{{{i}\alpha_{{1}}}}{\left|{z}_{{2}}\right|}{e}^{{{i}\alpha_{{2}}}}={\left|{z}_{{1}}{z}_{{2}}\right|}{e}^{{{i}{\left(\alpha_{{1}}+\alpha_{{2}}\right)}}},{\left({1}\right)}$$
so the argument of $$\displaystyle{z}_{{1}}{z}_{{2}}\ {i}{s}\ \alpha_{{1}}+\alpha_{{2}}.$$
Now define
$$\displaystyle\varphi:\mathbb{C}^{\ast}={\mathbb{R}_{{+}}^{{\ast}}}\times\top$$
with
$$\displaystyle\varphi{\left({z}\right)}=\varphi{\left({\left|{z}\right|}{e}^{{{i}\alpha}}\right)}={\left({\left|{z}\right|},{e}^{{{i}\alpha}}\right)}$$
It is well-defined since the representation $$\displaystyle{z}={\left|{z}\right|}{e}^{{{i}\alpha}}$$ is unique. It is also a homomorphism since
$$\displaystyle\varphi{\left({z}_{{1}}{z}_{{2}}\right)}=\varphi{\left({\left|{z}_{{1}}{z}_{{2}}\right|}{e}^{{{i}{\left(\alpha_{{1}}+\alpha_{{2}}\right)}}}={\left({\left|{z}_{{1}}{z}_{{2}}\right|},{e}^{{{i}{\left(\alpha_{{1}}+\alpha_{{2}}\right)}}}\right.}\right.},$$
and
$$\displaystyle\varphi{\left({z}_{{1}}\right)}\varphi{\left({z}_{{2}}\right)}={\left({\left|{z}_{{1}}\right|},{e}^{{{i}\alpha_{{1}}}}\right)}{\left({\left|{z}_{{2}}\right|}{e}^{{{i}\alpha_{{2}}}}\right)}={\left({\left|{z}_{{1}}{z}_{{2}}\right|},{e}^{{{i}{\left(\alpha_{{1}}+\alpha_{{2}}\right)}}},\right.}$$
hence
$$\displaystyle\varphi{\left({z}_{{1}}{z}_{{2}}\right)}=\varphi{\left({z}_{{1}}\right)}\varphi{\left({z}_{{2}}\right)}$$
Now we will prove that is injective. We know that varphi is injective if and only if ker $$\displaystyle\varphi={\left\lbrace{1}\right\rbrace}$$ (because 1 is a neutral element of $$\displaystyle\mathbb{C}^{\ast}$$. So, let z in ker $$\displaystyle\varphi$$.
Then
$$\displaystyle\varphi{\left({z}\right)}={\left({1},{1}\right)},$$
since (1, 1) is a neutral element of $$\displaystyle{\mathbb{R}_{{+}}^{{\ast}}}\times\top$$. Furthermore, this means that $$\displaystyle{\left({\left|{z}\right|},{e}^{{{i}\alpha}}\right)}={\left({1},{1}\right)}$$
Thus,
$$\displaystyle{\left|{z}\right|}={1}$$
and
$$\displaystyle{e}^{{{i}\alpha}}={1},$$
so
$$\displaystyle{z}={\left|{z}\right|}{e}^{{{i}\alpha}}={1}$$
We got that
ker $$\displaystyle\varphi\subseteq$$ {1}
Since clearly $$\displaystyle\varphi{\left({1}\right)}={\left({1},{1}\right)}$$, we get that {1} $$\displaystyle\subseteq{k}{e}{r}\varphi$$,
hence
ker $$\displaystyle\varphi={\left\lbrace{1}\right\rbrace}$$
Therefore, $$\displaystyle\varphi$$ is injective.
Finally, we will prove that $$\displaystyle\varphi$$ is surjective. But this is almost trivial, just take any $$\displaystyle{\left({x},{y}\right)}\in{\mathbb{R}_{+}^{{\ast}}}\times\top$$ notice that $$\displaystyle{x}{e}^{{{i}{y}}}\ne{0} \sin{{c}}{e}{x}\ne{0}{\quad\text{and}\quad}{e}^{{{i}{y}}}\ne{0},{\quad\text{and}\quad}\varphi{\left({x}{e}^{{{i}{y}}}\right)}={\left({x},{y}\right)},{s}{o}{\left({x},{y}\right)}\in{i}{m}\varphi$$. Thus $$\displaystyle\varphi$$ is surjective.Finally, $$\displaystyle\varphi$$ is an isomorphism of $$\displaystyle\mathbb{C}^{\ast}{\quad\text{and}\quad}{\mathbb{R}_{+}^{{\ast}}}\times\top$$, which completes the proof.Result recall that any complex number z can be written (uniquely!) as $$\displaystyle{z}={\left|{z}\right|}{e}^{{{i}\alpha}}$$