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.

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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.

Mathematics
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Is it not that the function f maps the coordination of a point to itself in unit circle?
show \[\phi(a+b)=\phi(a)\phi(b)\]
I know how to show it is a homomorphism. I want to make sure about the map

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yes it maps to the unit circle in the complex plane
Define: S := additive group R T as it is f: S --> T \(z=cis t\mapsto |z|\) right?
If it is so, it turns to easy, right? hehehe.... since |cis t| = 1.
@Zarkon Thank you so much.
yes...very easy

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