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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.

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  1. Loser66
    • one year ago
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    Is it not that the function f maps the coordination of a point to itself in unit circle?

  2. Zarkon
    • one year ago
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    show \[\phi(a+b)=\phi(a)\phi(b)\]

  3. Loser66
    • one year ago
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    I know how to show it is a homomorphism. I want to make sure about the map

  4. Zarkon
    • one year ago
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    yes it maps to the unit circle in the complex plane

  5. Loser66
    • one year ago
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    Define: S := additive group R T as it is f: S --> T \(z=cis t\mapsto |z|\) right?

  6. Loser66
    • one year ago
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    If it is so, it turns to easy, right? hehehe.... since |cis t| = 1.

  7. Loser66
    • one year ago
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    @Zarkon Thank you so much.

  8. Zarkon
    • one year ago
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    yes...very easy

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