Putnam 2005, Problem A-3

Let $p(z)$ be a polynomial of degree $n$ all of whose zeros have absolute value 1 in the complex plane. Put $g(z) = p(z)/z^{n/2}$. Show that all zeros of $g'(z) = 0$ have absolute value 1.

Solution:
Note that it is implicit in the problem that \(p\) is nonconstant, one may take any branch of the square root, and that \(z = 0\) should be ignored.

First solution: Write \(p(z) = c \prod_{j=1}^n (z - r_j)\), so that
\[
\frac{g'(z)}{g(z)} = \frac{1}{2z} \sum_{j=1}^n \frac{z + r_j}{z - r_j}.
\]
Now if \(z \neq r_j\) for all \(j\), then
\[
\frac{z + r_j}{z - r_j} = \frac{(z + r_j)(\bar{z} - \bar{r}_j)}{|z - r_j|^2} = \frac{|z|^2 - 1 + 2\text{Im}(\bar{z}r_j)}{|z - r_j|^2},
\]
and so
\[
\text{Re} \frac{zg'(z)}{g(z)} = \frac{|z|^2 - 1}{2} \left( \sum_{j} \frac{1}{|z - r_j|^2} \right).
\]
Since the quantity in parentheses is positive, \(g'(z)/g(z)\) can be 0 only if \(|z| = 1\). If on the other hand \(z = r_j\) for some \(j\), then \(|z| = 1\) anyway.

Second solution: Write \(p(z) = c \prod_{j=1}^n (z - r_j)\), so that
\[
\frac{g'(z)}{g(z)} = \sum_{j=1}^n \left( \frac{1}{z - r_j} - \frac{1}{2z} \right).
\]
We first check that \(g'(z) \neq 0\) whenever \(z\) is real and \(z > 1\). In this case, for \(r_j = e^{i\theta_j}\), we have \(z - r_j = (z - \cos(\theta_j)) + \sin(\theta_j)i\), so the real part of \(\frac{1}{z - r_j} - \frac{1}{2z}\) is
\[
\frac{z - \cos(\theta_j)}{z^2 - 2z\cos(\theta_j) + 1} - \frac{1}{2z} = \frac{z^2 - 1}{2z(z^2 - 2z\cos(\theta_j) + 1)} > 0.
\]
Hence \(g'(z)/g(z)\) has positive real part, so \(g'(z)/g(z)\) and hence \(g(z)\) are nonzero.

Applying the same argument after replacing \(p(z)\) by \(p(e^{i\theta}z)\), we deduce that \(g'\) cannot have any roots outside the unit circle. Applying the same argument after replacing \(p(z)\) by \(z^n p(1/z)\), we also deduce that \(g'\) cannot have any roots inside the unit circle. Hence all roots of \(g'\) have absolute value 1, as desired.

Third solution: Write \(p(z) = c \prod_{j=1}^n (z - r_j)\) and put \(r_j = e^{2i\theta_j}\). Note that \(g(e^{2i\theta})\) is equal to a nonzero constant times
\[
h(\theta) = \prod_{j=1}^n \frac{e^{i(\theta + \theta_j)} - e^{-i(\theta + \theta_j)}}{2i} = \prod_{j=1}^n \sin(\theta + \theta_j).
\]
Since \(h\) has at least \(2n\) roots (counting multiplicity) in the interval \([0,2\pi)\), \(h'\) does also by repeated application of Rolle's theorem. Since \(g'(e^{2i\theta}) = 2ie^{2i\theta} h'(\theta)\), \(g'(z^2)\) has at least \(2n\) roots on the unit circle. Since \(g'(z^2)\) is equal to \(z^{-n-1}\) times a polynomial of degree \(2n\), \(g'(z)\) has all roots on the unit circle, as then does \(g'(z)\).

Remarks: The second solution imitates the proof of the Gauss-Lucas theorem: the roots of the derivative of a complex polynomial lie in the convex hull of the roots of the original polynomial. The second solution is close to problem B3 from the 2000 Putnam. A hybrid between the first and third solutions is to check that on the unit circle, \(\text{Re}(zg'(z)/g(z)) = 0\) while between any two roots of \(p\), \(\text{Im}(zg'(z)/g(z))\) runs from \(+\infty\) to \(-\infty\) and so must have a zero crossing. (This only works when \(p\) has distinct roots, but the general case follows by the continuity of the roots of a polynomial as functions of the coefficients.) One can also construct a solution using Rouché's theorem.