Question

Show that \(\phi(t) = cis t\) is a group homomorphism of the additive group\(\mathbb R\) onto the multiplicative group T={z : |z|=1}\)
Please help.

Answer 1

Is it not that the function f maps the coordination of a point to itself in unit circle?

Answer 2

show
\[\phi(a+b)=\phi(a)\phi(b)\]

Answer 3

I know how to show it is a homomorphism. I want to make sure about the map

Answer 4

yes it maps to the unit circle in the complex plane

Answer 5

Define:
S := additive group R
T as it is
f: S --> T
\(z=cis t\mapsto |z|\) right?

Answer 6

If it is so, it turns to easy, right? hehehe.... since |cis t| = 1.

Answer 7

@Zarkon Thank you so much.

Answer 8

yes...very easy