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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.
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.

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Is it not that the function f maps the coordination of a point to itself in unit circle?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1show \[\phi(a+b)=\phi(a)\phi(b)\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I know how to show it is a homomorphism. I want to make sure about the map

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1yes it maps to the unit circle in the complex plane

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Define: S := additive group R T as it is f: S > T \(z=cis t\mapsto z\) right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0If it is so, it turns to easy, right? hehehe.... since cis t = 1.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@Zarkon Thank you so much.
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