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

Step 1: Plan the count using Rouche's theorem.
Let \( f(z) = z^6+5z^3+4z+11 \). To count zeros in the annulus \( 1 < |z| < 3 \), first count zeros inside \( |z| < 3 \), then count zeros inside \( |z| < 1 \), and subtract. Rouche's theorem says that if \( |g(z)| < |h(z)| \) on a circle, then \( h+g \) and \( h \) have the same number of zeros inside that circle.

Step 2: Count zeros inside \( |z| = 3 \).
Split \( f(z) = z^6 + (5z^3+4z+11) \), taking \( h(z) = z^6 \) as the dominant piece. On \( |z| = 3 \):
\[ |z^6| = 3^6 = 729 \]
\[ |5z^3+4z+11| \le 5|z|^3 + 4|z| + 11 = 5(27) + 4(3) + 11 = 135+12+11 = 158 \]
Since \( 158 < 729 \), Rouche's theorem applies with \( h(z)=z^6 \) and \( g(z)=5z^3+4z+11 \), so \( f \) has the same number of zeros inside \( |z|<3 \) as \( z^6 \) does, which is \( 6 \) (all at the origin, counted with multiplicity). So \( f \) has \( 6 \) zeros inside \( |z| < 3 \), which accounts for all \( 6 \) roots of this degree-\( 6 \) polynomial.

Step 3: Count zeros inside \( |z| = 1 \).
Now split \( f(z) = 11 + (z^6+5z^3+4z) \), taking the constant \( h(z) = 11 \) as dominant this time. On \( |z| = 1 \):
\[ |z^6+5z^3+4z| \le |z|^6 + 5|z|^3 + 4|z| = 1+5+4 = 10 \]
Since \( 10 < 11 = |h(z)| \), Rouche's theorem again applies, so \( f \) has the same number of zeros inside \( |z| < 1 \) as the constant function \( 11 \) does. A nonzero constant has \( 0 \) zeros, so \( f \) has \( 0 \) zeros inside \( |z| < 1 \).

Step 4: Subtract to get the count in the annulus.
Zeros in \( 1 < |z| < 3 \) equals (zeros inside \( |z|<3 \)) minus (zeros inside \( |z| \le 1 \)), which is \( 6 - 0 = 6 \). Neither bound above was tight (\( 158<729 \) and \( 10<11 \)), so no zero lies exactly on \( |z|=1 \) or \( |z|=3 \), and the count of \( 6 \) zeros strictly inside the annulus is exact.

Final Answer:
The number of zeros of \( z^6+5z^3+4z+11 \) in the annulus \( 1<|z|<3 \) is \( 6 \). \[ \boxed{6} \]