Find the number of roots of \[ z^7-4z^3-11=0 \] that lie between the circles \[ |z|=1 \qquad\text{and}\qquad |z|=2. \]
Solution
Let \[ p(z)=z^7-4z^3-11. \]
1. The polynomial \(p\) has no zero in the closed unit disk.
On \(\abs z=1\), \[ \abs{z^7-4z^3} \leq \abs{z}^7+4\abs{z}^3 = 5 < 11 = \abs{-11}. \] By Rouché’s theorem, \(p(z)=-11+(z^7-4z^3)\) and the constant function \(-11\) have the same number of zeros in \(\abs z<1\), namely none. The strict inequality also shows that \(p\) has no zero on \(\abs z=1\).
2. The polynomial \(p\) has seven zeros in the disk \(\abs z<2\),counted with multiplicity.
On \(\abs z=2\), \[ \abs{-4z^3-11} \leq 4\abs z^3+11 = 43 < 128 = \abs{z^7}. \] By Rouché’s theorem, \(p(z)=z^7+(-4z^3-11)\) and \(z^7\) have the same number of zeros in \(\abs z<2\). Thus \(p\) has seven zeros there, counted with multiplicity. Again, the strict inequality excludes zeros on \(\abs z=2\).
3. The number of roots in\[ 1<\abs z<2 \]is\[ \boxed{7}, \]counted with multiplicity.
Step 2 counts seven zeros inside the outer circle, while step 1 counts zero zeros on or inside the inner circle. Therefore all seven zeros inside \(\abs z<2\) lie in the stated annulus.
4. Q.E.D.
Step 3 gives the requested root count.