How many roots does the equation \(z ^ { 7 } - 4 z ^ { 3 } - 1 = 0\) have in the open disk \(| z | < 1 \ \mathrm { ? }\)
Solution
Set \[ p(z)=z^7-4z^3-1. \]
1. On the unit circle,\[ \abs{z^7-1} < \abs{-4z^3}. \]
If \(\abs{z}=1\), then \[ \abs{z^7-1} \leq \abs{z}^7+1 = 2, \] while \[ \abs{-4z^3} = 4\abs{z}^3 = 4. \] Hence the strict inequality holds.
2. The polynomial \(p\) has exactly three zeros in the open unit disk,counting multiplicity.
Write \[ p(z)=-4z^3+(z^7-1). \] By step 1, Rouché’s theorem implies that \(p\) and \(-4z^3\) have the same number of zeros in \(\abs{z}<1\), counting multiplicity. The polynomial \(-4z^3\) has exactly three zeros there, all at \(z=0\) counted with multiplicity.
3. The answer is\[ \boxed{3}. \]
This is step 2.
4. Q.E.D.
Step 3 gives the requested number of roots.