I'm studying for my PhD prelim exam in complex analysis, and I ran into this example problem.

Show that the polynomial $$p(z)=z^{47} − z^{23} + 2z^{11} − z^5 + 4z^2 + 1$$ has at least one root inside the unit disk $\{\lvert z\rvert <1\}$.

My initial thought was to try to use Rouché's theorem, but none of the terms dominate all of the other terms on the unit circle, so I can't apply Rouché in an obvious way.

Alternatively, we know that the product of all of the roots must equal 1. So either all of the roots have unit modulus or there is at least one root inside the unit circle. But I can't figure out how to show that at least one root does not have unit modulus.

What am I missing?


1 Answer 1


Elementary solution We exploit the insight that the product of all of the roots (and hence the product of the moduli of the roots) is $1$:

Since $p$ has odd degree and real coefficients, it has at least one real root. But substituting gives that $p(-1) = 4$ and $p(1) = 6$, so $p$ has a real root that in particular does not have unit modulus. Thus $p$ must have at least one root of modulus $> 1$ and one root of modulus $< 1$.

Rouché's Theorem Alternatively, one can use Rouché's Theorem (which was probably the intent of the problem author) to prove the stronger result that $p$ has a root in some smaller disc $\{|z| < R\}$. For example, some easy arithmetic and the Triangle Inequality shows that $$|4 z^2| = 4 |z|^2 > |z|^{47} + |z|^{23} + 2 |z|^{11} + |z|^5 + 1 \geq |p(z) - 4 z^2|$$ on the circle $$\left\{|z| = \frac{1}{\sqrt{2}}\right\},$$ so we may apply Rouché's theorem to the disc of radius $R = \frac{1}{\sqrt{2}}$ and so conclude the desired result, though any radius $$0.509 \leq R \leq 0.961$$ will do.

In fact, since $4 z^2$ has two roots, counting multiplicity, in any such disc, so does $p(z)$.

  • $\begingroup$ Thanks! The solution is definitely quite a bit simpler than I thought it was going to be. $\endgroup$ Dec 14, 2014 at 18:50
  • $\begingroup$ You're welcome, I'm glad you found it useful. If you wanted to use Rouch\'{e}'s Theorem, you could use it to prove the stronger statement that there is a root in $\left\{|z| < \frac{1}{2}\right\}$, as the term $4 z^2$ dominates in the appropriate way on its boundary. $\endgroup$ Dec 14, 2014 at 19:07
  • $\begingroup$ I tried that. Unfortunately it doesn't work, since we only get $\lvert 4z^2\rvert=1$ on $\lvert z\rvert = \frac{1}{2}$. And this doesn't dominate the rest of the polynomial on this circle. $\endgroup$ Dec 14, 2014 at 20:07
  • 1
    $\begingroup$ Oops, rather, try the disc $\left\{|z| < \frac{1}{\sqrt{2}}\right\}$. On the boundary circle, $|4 z^2| = 2$ but the other terms have total modulus $< \frac{3}{2}$. (In fact, any disc of radius $0.509 \leq r \leq 0.961$ does the trick using the naive modulus estimate.) $\endgroup$ Dec 14, 2014 at 20:25
  • $\begingroup$ Aha! Before I posted here on stackexchange, I tried finding a radius between $\frac{1}{2}$ and $1$ that would work for Rouches theorem, but got stuck. That's more like the solution I was looking for. Thanks again! $\endgroup$ Dec 14, 2014 at 20:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .