If $p(z)$ is an injective polynomial $\Longrightarrow$ $p(z)=az+b$

Question

If $p(z)$ is an injective polynomial, how to prove that $p(z)=az+b$ with $a\neq 0$.

$p(z)\in\mathbb{C}[z]$.

Any hint would be appreciated.

Answer 1

If $p(z) \in \Bbb C[z]$ is injective, we clearly cannot have $\deg p(z) = 0$, since then $p(z)$ is a constant, $p(z) = c \in \Bbb C$ for all $z \in \Bbb C$; not injective! If $\deg p(z) = n \ge 2$, then $p(z)$ has $n$ zeroes when they are counted with their multiplicities. If there are two distinct roots $x \ne y$, then $p(x) = p(y) = 0$; $p(z)$ is not injective. If there is one zero $x$ of multiplicity $n$, then $p(z) = c(z - x)^n$ for some nonzero $c \in \Bbb C$. But $c(z - x)^n$ maps $n$ values to any $y \ne x$, viz. setting $\frac{y}{c} = re^{i\theta}$ with $0 \le \theta < 2\pi$, $p(x + r^{1/n}e^{i(\theta/n)}e^{i(2k\pi/n)}) = y$ for $0 \le k < n$, as is easily seen by direct computation. So such $p(z)$ cannot be injective either; thus we must have $n = 1$ and $p(z)$ is linear. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 2

$p(z)=a$ doesn't work so consider $p(z)=Q(z)+b$ where $Q(z)=\sum_{j=1}^n a_jz^j$ with $n\geq 1$ and $a_n\neq 0$. We want to show that $p(z)$ is not injective if $n>1$. Either there is $z'\neq 0$ such that $Q(z')=0$ in which case $p(0)=p(z')=b$, or $Q(z)=a_nz^n$. In this case $p(z_1)=p(z_2)=b+a_n$ for any $z_1$ and $z_2$ that are distinct $n$-th roots of unity.

Answer 3

The proof https://math.stackexchange.com/a/35471/27978 shows that if an analytic function $f$ satisfies $f'(z_0) = 0$, then $f$ is not injective.

Hence we have $p'(z) \neq 0$ for all $z$. Since $p'$ is a polynomial, the only way this can happen is if it is a non-zero constant. Hence $p(z) = p(0)+p'(0)z$.

Answer 4

Fix $p\in \mathbb{C}[X]$ with $\deg p > 1$. Let $z_1, \dots, z_r$ denote the zeros of $p'$, and choose $w\in\mathbb{C}$ with $w\not = p(z_i)$ for each $i$. The polynomial $q(z) = p(z) - w$ then has no common zeros with $q' = p'$. Thus the preimage $q^{-1}(0) = p^{-1}(w)$ contains exactly $\deg q = \deg p > 1$ points, and so $p$ is not injective. (If the preceding sentence isn't clear, try computing $f'(z_i)$ for $f(z) = (z - z_1) \cdots (z - z_n)$, being careful about what happens when some of the $z_i$ coincide.)

Answer 5

Suppose $p$ is injective (in particular, $p$ is not constant). Let $n=\partial p$ be the degree of $p$ and $\lambda_1,\ldots,\lambda_n$ its roots, so that $p(z)=a(z-\lambda_1)\cdots(z-\lambda_n)$ for some $a\in\mathbb{C}\setminus\left\{0\right\}$.

Since $p(\lambda_1)=\cdots=p(\lambda_n)=0$, then, by injectivity of $p$, $\lambda_1=\cdots=\lambda_n$, that is, $p(z)=a(z-\lambda)^n$, where $\lambda=\lambda_1$. Let's show that $n=1$.

Suppose otherwise, that is, $n\geq 2$. Let $x$ and $x'$ be two distinct $n$th roots of unity. Then $p(\lambda+x)=a=p(\lambda+x')$, contradicting injectiveness of $p$. Therefore, $n=1$, and $p(z)=a(z-\lambda)=az-a\lambda$. Simply take $b=-a\lambda$ to obtain the result.

(PS. If you don't like proofs by contradiction, you can use the same idea to have a direct, but a little longer, proof: Let $x=\cos(2\pi/n)+i\sin(2\pi/n)$ (the usual $n$th root of unity). Then $p(x+\lambda)=a=p(1+\lambda)$. Since $p$ is injective, then $x=1$, so $\cos(2\pi/n)=1$. But also, $0<2\pi/n\leq2\pi$, and the only point of $(0,2\pi]$ in which $\cos$ attains $1$ is $2\pi$, so $2\pi/n=2\pi$, hence $n=1$.)