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Prove that $z^k+kz$ is 1-1 on the unit disk for $k \geq 1$ and $k \in \mathbb N$.

My proof: Take $a \in \mathbb C$, and consider $g(z) = z^k + kz -a$. Then if it has more than one roots at $z$, its derivative should vanish at that point too. So $k z^{k-1} + k = 0$, which says that multiple rootS can only appear on the boundary of the unit disk. So we are done.

I do not really use the fact that $k \leq 1$, which is clearly suggesting Rouche's theorem. It might be the problem is asking about the closed unit disk. Is my proof correct?

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  • $\begingroup$ Is $k$ real...? $\endgroup$
    – Nosrati
    Commented Sep 18, 2018 at 16:32
  • $\begingroup$ What is $z^k$ on the unit disk if $k$ is not a positive integer? $\endgroup$
    – Did
    Commented Sep 18, 2018 at 16:33

2 Answers 2

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$$z_1^k+kz_1=z_1^k+kz_1$$ shows $$(z_1-z_2)(z_1^{k-1}+z_1^{k-2}z_2+\cdots+z_1z_2^{k-2}+z_2^{k-1}+k)=0$$ but with $|z_1|<1$ and $|z_2|<1$ $$|z_1^{k-1}+z_1^{k-2}z_2+\cdots+z_1z_2^{k-2}+z_2^{k-1}+k|\geq k-|z_1|^{k-1}-|z_1|^{k-2}|z_2|-\cdots-|z_1||z_2|^{k-2}-|z_2|^{k-1}>0$$

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Your proof is wrong. The function can fail to be one-to-one by $z^k+kz-a$ having two distinct zeros in the unit disk, not just by having a zero of multiplicity $> 1$.

But the problem is a bit strange, because if $k$ is not a natural number $z^k +k z$ is not analytic on the unit disk.

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    $\begingroup$ Rather a comment? $\endgroup$
    – Did
    Commented Sep 18, 2018 at 16:32
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    $\begingroup$ The question asked "Is my proof correct?". I answered that. $\endgroup$ Commented Sep 18, 2018 at 16:33
  • $\begingroup$ Sorry I made a mistake in the condition. $\endgroup$
    – zach
    Commented Sep 18, 2018 at 16:33
  • $\begingroup$ What should be the right approach? I am thinking of Rouche's theorem, but with the extra $a$ there, it seems hard to apply Rouche. $\endgroup$
    – zach
    Commented Sep 18, 2018 at 16:35
  • $\begingroup$ @RobertIsrael ^^^. $\endgroup$
    – Did
    Commented Sep 18, 2018 at 16:37

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