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I'm trying to see if the function:

$$z \mapsto z^n+\exp(ia) \cdot nz$$

is an injective function at the open unit circle.
Please help.

  • 0
    What does the '*' mean?2012-11-24
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    I'm guessing it means multiplication, @copper.hat .2012-11-24
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    It means multiplication :)2012-11-24
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    @DonAntonio: Thanks, I would have thought so, but then I would have expected $i*a$ and $n*z$ as well?2012-11-24
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    The function satisfies the necessary condition from De Branges' Theorem.2012-11-24

1 Answers 1

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$$z^n+nze^{ia}=w^n+nwe^{ia}\Longrightarrow (z-w)(z^{n-1}+z^{n-2}w+...+zw^{n-2}+w^{n-1})=-ne^{ia}(z-w)$$

If $\,z\neq w\,$ then $\,z^{n-1}+z^{n-2}w+...+zw^{n-2}+w^{n-1}=-ne^{ia}$

But, assuming $\,a\in\Bbb R\,$, we get that the RHS's module is $\,n\,$, whereas the LHS's module is $\,|z^{n-1}+z^{n-2}w+...+zw^{n-2}+w^{n-1}|<1+1+...+1 = n$