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

e1s2kat26 2021-03-04 Answered

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.

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Expert Answer

Bentley Leach
Answered 2021-03-05 Author has 109 answers

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 (z0))
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

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