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Suppose f(z) is a polynomial such that its derivatives are non-zero for all $|z| <1$. Is the restriction of $f$ to $|z|<1$ 1 to 1?

I know that $f$ must be locally 1 to 1. It is obvious that $f$ is 1-1 for polynomials of order 1. The case of order 2 follows from Gauss-Lucas Thm.

I am stuck on how to prove the general case.

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2 Answers 2

up vote 6 down vote accepted

It's not true in general... consider $f(z) = (z-1)^n - 1$. Then its derivatives have zeroes only at $z = 1$, but it takes the value zero at any $z = 1 + e^{2\pi i / n}$. Since these are spaced evenly around the circle $|z - 1|= 1$, for large $n$ many of them will lie inside the unit disk.

If you want an example where the derivatives have zeroes only outside the ${\it closed}$ unit disk, then you can take $f(z) = (z-(1+\epsilon))^n - 1$ for small $\epsilon > 0$.

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Zarrax has already given a great example. Here's a way to see that there are polynomial examples from the fact that there are holomorphic examples. The motivation for this is that the non-polynomial example $g(z)=e^{2\pi i z}$ came to mind when I read your question.

Let $g$ be holomorphic in a neighborhood of the closed disk, not injective on the open disk, but with nonvanishing derivative on the closed disk. Let $(p_n)_n$ be the sequence of partial sums of the Taylor series of $g$ centered at $0$, so $p_n\to g$ and $p_n'\to g'$ uniformly on a (smaller) neighborhood of the closed disk.

Since $g'$ is nonvanishing on the closed disk, it has a positive minimum modulus $m$ there, which by uniform convergence means that $p_n'$ eventually has modulus greater than $m/2$, and in particular is eventually nonvanishing on the closed disk. Take $a\neq b$ in the open unit disk such that $g(a)=g(b)$. Applying Hurwitz's theorem to the sequence of functions $p_n(z)-g(a)$ on small disjoint disks centered at $a$ and $b$ shows that $p_n$ eventually takes on the value $g(a)$ at more than one point in the open unit disk. Thus, $p_n$ is eventually a counterexample.

In fact, using WolframAlpha, it looks like $f(z)=\displaystyle{\sum_{k=0}^{25}\frac{(2\pi iz)^k}{k!}}$ gives an example. It allegedly takes on the value $-1$ at points very close to $\pm\frac{1}{2}$, and all $24$ of the zeros of its derivative are outside the closed disk.

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The following reference on zeros of partial sums of the exponential function gives results that can be used to proving the claims of the zeros of the derivative WolframAlpha gave me: math.washington.edu/~morrow/336_09/papers/Ian.pdf –  Jonas Meyer Feb 2 '11 at 7:17
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