I'm trying to solve the following problem.

Any polynomial function injective on the upper half plane has degree less than or equal to 2.

This seems intuitively true, since polynomial of degree $n$ will approximately be an $n$ fold covering map. However I have no idea how to prove it explictely. I tried to show that polynomial of degree > 2 is not injective by finding two points having the same value, but it didn't work well. Any idea for this one?


2 Answers 2


Assume $P$ is a degree $3$ or higher polynomial; wlog we can assume $P$ monic since dividing by a non zero constant doesn't change injectivity, so $P(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_0$ and in particular there is $R_P>0$ st for $|z| \ge R_P$ we have $|z^n| >|P(z)-z^n|$

Now since $n \ge 3$ the equation $z^n=e^{in\pi/4}$ has at least two roots $e^{\pi i/4}, e^{\pi i/4+2\pi i/n}$ in the upper half plane since $1/4+2/n \le 1/4+2/3 <1$ and let those two roots be $e^{i\alpha}, e^{i\beta}$ with $\sin \alpha, \sin \beta >\delta >0$ fixed.

Hence if we consider the rectangle $G$ bounded by $x= \pm 2R, y=R\delta, y=2R$ where we take $R \delta >R_P$, then $R \sin \alpha >R\delta, R\sin \alpha < 2R$ and same for $R \sin \beta$ while clearly $|R \cos \alpha| <2R, |R \cos \beta| <2R$, so both $Re^{i\alpha}, Re^{i\beta}$ are inside $G$; but we have that $|P(z)-z^n| <|z|^n$ on the boundary of $G$ since $w \in \bar G$ implies $|w|>R_P$ by our choices as $\Im w \ge R \delta >R_P$

Hence by Rouche $P(z)=R^ne^{in\pi/4}$ and $z^n=R^ne^{in\pi/4}$ have same number of roots in $G$, while the latter equation has at least the two roots $Re^{i\alpha}, Re^{i\beta}$, so $P$ cannot be injective in $G$


Let $$ p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0 $$ be a monic polynomial of degree $n \ge 3$. We want to show that $p$ is not injective in the upper half-plane.

(The idea is that $p(z) \sim z^n$ for sufficiently large $z$, so that $p$ takes the same values at points which roughly differ by an argument of $2\pi/n$, and some large values must be taken twice in the upper half-plane.)

The function $$ f(z) = \frac{1}{p(1/z)} = \frac{z^n}{1 + a_{n-1}z + \ldots + a_0z^n} $$ is holomorphic in a sufficiently small disk $B_r(0)$, with an $n$-fold zero at the origin. It follows that $$ f(z) = h(z)^n $$ where $h$ is holomorphic and injective in $B_r(0)$ with $h(0) = 0$ and $h'(0) = 1$. The preimage of a small segment $(-i \epsilon, 0)$ on the negative imaginary axis are $n$ curves $\gamma_k$, emanating from the origin with an argument of $$ \alpha_k = -\frac{\pi}{2n} - k \frac{2 \pi}{n}, \, k = 0, \ldots, n-1 \, . $$ Since $n \ge 3$, both $\alpha_0$ and $\alpha_1$ are in the range $(-\pi, 0)$, so that the initial parts of both $\gamma_0$ and $\gamma_1$ are in the lower half-plane.

So $f$ takes values on the segment $(-i \epsilon, 0)$ at least twice in the lower half-plane, which means that $p$ takes some values at least twice in the upper half-plane.

This shows that a polynomial of degree $\ge 3$ can not be injective in the upper half-plane.


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