Question

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

Vectors and spaces
ANSWERED
asked 2021-03-04

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.

Answers (1)

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 \(\displaystyle{z}={\left|{z}\right|}{e}^{{{i}\alpha}}\)

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