## Loser66 one year ago 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.

1. Loser66

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

2. Zarkon

show $\phi(a+b)=\phi(a)\phi(b)$

3. Loser66

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

4. Zarkon

yes it maps to the unit circle in the complex plane

5. Loser66

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

6. Loser66

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

7. Loser66

@Zarkon Thank you so much.

8. Zarkon

yes...very easy