Show that C∗=R+∗×T,where C∗ is the multiplicative group of non-zero complex numbers, T is the group of complex numbers of modulus equal to 1, R+∗ is the multiplicative group of positive real numbers.

Question

Show that C∗=R+∗×T,where C∗ is the multiplicative group of non-zero complex numbers, T is the group of complex numbers of modulus equal to 1, R+∗ is the multiplicative group of positive real numbers.

Bentley Leach · Accepted Answer

Recall that each nonzero complex number z can be written (uniquely!) in the form z=|z|eiα. where |z| is the modulus of z, and α∈[0,2π⟩ is the argument of z. Also, eiα is an element of TT, and |z| is a real positive number (positive since (z≠0)) Furthermore, ≤tz1andz2 be two nonzero complex numbers and z1=|z1|eiα1 z2=|z2|eiα2 Then z1z2=|z1|eiα1|z2|eiα2=|z1z2|ei(α1+α2),(1) so the argument of z1z2 is α1+α2. Now define φ:C∗=R+∗×⊤ with φ(z)=φ(|z|eiα)=(|z|,eiα) It is well-defined since the representation z=|z|eiα is unique. It is also a homomorphism since φ(z1z2)=φ(|z1z2|ei(α1+α2)=(|z1z2|,ei(α1+α2), and φ(z1)φ(z2)=(|z1|,eiα1)(|z2|eiα2)=(|z1z2|,ei(α1+α2), hence φ(z1z2)=φ(z1)φ(z2) Now we will prove that is injective. We know that varphi is injective if and only if ker φ={1} (because 1 is a neutral element of C∗. So, let z in ker φ. Then φ(z)=(1,1), since (1, 1) is a neutral element of R+∗×⊤. Furthermore, this means that

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

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