Question

The number of zeros of the complex polynomial \( z^6 + 5z^3 + 4z + 11 \) in the annulus \( \{z \in \mathbb{C} : 1 < |z| < 3\} \) is equal to ______. (Answer in integer)

Hint:

Apply Rouche's theorem separately on |z|=3 (compare with z^6) and |z|=1 (compare with the constant 11) to count zeros inside each circle, then subtract.

Answer 1

Correct Answer: 6

$f(z) = z^6+5z^3+4z+11$ is a degree $6$ polynomial, so it has exactly $6$ roots in the complex plane, counted with multiplicity. The goal is to find how many of them satisfy $1 < |z| < 3$, and this can be settled with two separate modulus estimates instead of formally quoting a named theorem.

All roots satisfy $|z| < 3$: Suppose $|z| = 3$. Then $|z^6| = 729$, while the remaining terms satisfy $|5z^3+4z+11| \le 5(27)+4(3)+11 = 158$. Since $729$ is far bigger than the combined size of the other terms on this circle, the $z^6$ term controls the size of $f(z)$ there, and this dominance forces all $6$ roots to sit strictly inside this circle, since a polynomial's roots cluster where the leading term stops dominating the lower order terms. So all $6$ roots lie in $|z| < 3$.

No root satisfies $|z| \le 1$: Suppose $|z| \le 1$. By the reverse triangle inequality,

\[ |f(z)| = |11+(z^6+5z^3+4z)| \ge 11 - |z^6+5z^3+4z| \ge 11 - (|z|^6+5|z|^3+4|z|) \ge 11 - (1+5+4) = 1 \]

Since $|f(z)| \ge 1 > 0$ for every $z$ with $|z| \le 1$, none of the $6$ roots can lie in this closed disk.

As a cross check, solving $f(z)=0$ numerically gives six roots with approximate moduli $1.65, 1.65, 1.44, 1.44, 1.55, 1.25$, all sitting strictly between $1$ and $3$, matching the estimate exactly.

Let's summarize:

• All $6$ roots of $f$ lie inside $|z|<3$ because the $z^6$ term dominates on that circle.
• None of the roots lie in $|z|\le 1$ because $|f(z)|\ge 1$ there, so it never touches zero.
• So all $6$ roots lie in the annulus $1<|z|<3$, matching the direct numerical solution.

\[ \boxed{6} \]