# Complex injective function

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.

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What does the '*' mean? –  copper.hat Nov 24 '12 at 14:57
I'm guessing it means multiplication, @copper.hat . –  DonAntonio Nov 24 '12 at 14:59
It means multiplication :) –  Ran Kashtan Nov 24 '12 at 15:00
@DonAntonio: Thanks, I would have thought so, but then I would have expected $i*a$ and $n*z$ as well? –  copper.hat Nov 24 '12 at 15:00
The function satisfies the necessary condition from De Branges' Theorem. –  Hans Engler Nov 24 '12 at 15:04

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