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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
    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 $\,|z^{n-1}+z^{n-2}w+...+zw^{n-2}+w^{n-1}|<1+1+...+1 = n$