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

Show that is the multiplicative group of non-zero complex numbers, T is the group of complex numbers of modulus equal to 1, ${R}_{+}^{\ast }$ is the multiplicative group of positive real numbers.

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Bentley Leach

Recall that each nonzero complex number z can be written (uniquely!) in the form
$z=|z|{e}^{i\alpha }.$
where $|z|$ is the modulus of z, and $\alpha \in \left[0,2\pi ⟩$ is the argument of z. Also,
${e}^{i\alpha }$ is an element of TT, and $|z|$ is a real positive number (positive since $\left(z\ne 0\right)$)
Furthermore, $\le t{z}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{2}$ be two nonzero complex numbers and
${z}_{1}=|{z}_{1}|{e}^{i{\alpha }_{1}}$
${z}_{2}=|{z}_{2}|{e}^{i{\alpha }_{2}}$
Then
${z}_{1}{z}_{2}=|{z}_{1}|{e}^{i{\alpha }_{1}}|{z}_{2}|{e}^{i{\alpha }_{2}}=|{z}_{1}{z}_{2}|{e}^{i\left({\alpha }_{1}+{\alpha }_{2}\right)},\left(1\right)$
so the argument of
Now define
$\phi :{\mathbb{C}}^{\ast }={\mathbb{R}}_{+}^{\ast }×\mathrm{\top }$
with
$\phi \left(z\right)=\phi \left(|z|{e}^{i\alpha }\right)=\left(|z|,{e}^{i\alpha }\right)$
It is well-defined since the representation $z=|z|{e}^{i\alpha }$ is unique. It is also a homomorphism since
$\phi \left({z}_{1}{z}_{2}\right)=\phi \left(|{z}_{1}{z}_{2}|{e}^{i\left({\alpha }_{1}+{\alpha }_{2}\right)}=\left(|{z}_{1}{z}_{2}|,{e}^{i\left({\alpha }_{1}+{\alpha }_{2}\right)},$
and
$\phi \left({z}_{1}\right)\phi \left({z}_{2}\right)=\left(|{z}_{1}|,{e}^{i{\alpha }_{1}}\right)\left(|{z}_{2}|{e}^{i{\alpha }_{2}}\right)=\left(|{z}_{1}{z}_{2}|,{e}^{i\left({\alpha }_{1}+{\alpha }_{2}\right)},$
hence
$\phi \left({z}_{1}{z}_{2}\right)=\phi \left({z}_{1}\right)\phi \left({z}_{2}\right)$
Now we will prove that is injective. We know that varphi is injective if and only if ker $\phi =\left\{1\right\}$ (because 1 is a neutral element of ${\mathbb{C}}^{\ast }$. So, let z in ker $\phi$.
Then
$\phi \left(z\right)=\left(1,1\right),$
since (1, 1) is a neutral element of ${\mathbb{R}}_{+}^{\ast }×\mathrm{\top }$. Furthermore, this means that