How many roots of the equation \(z^4 - 5z^3 + z - 2 = 0\) lie in the disk \(|z| < 1\)?
Solution
Set \[ F(z)\coloneqq z^4-5z^3+z-2 \] and \[ G(z)\coloneqq-5z^3. \]
1. On the unit circle,\[ \abs{F(z)-G(z)}<\abs{G(z)}. \]
If \(\abs z=1\), then \[ \begin{aligned} \abs{F(z)-G(z)} &= \abs{z^4+z-2}\\ &\leq \abs z^4+\abs z+2\\ &= 4 < 5 = \abs{-5z^3} = \abs{G(z)}. \end{aligned} \]
2. The polynomials \(F\) and \(G\) have the same number of zeros in\(\abs z<1\), counted with multiplicity.
Both functions are holomorphic on and inside the unit circle, and step 1 is the strict inequality required by Rouché’s theorem. The conclusion follows directly from that theorem.
3. The equation has\[ \boxed{3} \]roots in the disk \(\abs z<1\), counted with multiplicity.
The polynomial \[ G(z)=-5z^3 \] has exactly one zero, \(z=0\), of multiplicity \(3\). Step 2 transfers this count to \(F\).
4. Q.E.D.
Step 3 gives the requested number of roots.