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?
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 $p(z)$.