# Show that $z^k+kz$ is $1-1$ on the unit disk, for every natural number $k$

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?

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

$$z_1^k+kz_1=z_1^k+kz_1$$ 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 $|z_2|<1$ $$|z_1^{k-1}+z_1^{k-2}z_2+\cdots+z_1z_2^{k-2}+z_2^{k-1}+k|\geq k-|z_1|^{k-1}-|z_1|^{k-2}|z_2|-\cdots-|z_1||z_2|^{k-2}-|z_2|^{k-1}>0$$

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.

• Rather a comment?
– Did
Commented Sep 18, 2018 at 16:32
• The question asked "Is my proof correct?". I answered that. Commented Sep 18, 2018 at 16:33
• Sorry I made a mistake in the condition.
– zach
Commented Sep 18, 2018 at 16:33
• 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.
– zach
Commented Sep 18, 2018 at 16:35
• @RobertIsrael ^^^.
– Did
Commented Sep 18, 2018 at 16:37