Let \(0<r<1<R\). Show that for all sufficiently small \(\varepsilon>0\), the polynomial \[ p(z)=\varepsilon z^7+z^2+1 \] has exactly five roots, counted with multiplicity, in the annulus \[ r\varepsilon^{-1/5}<|z|<R\varepsilon^{-1/5}. \]
The source prints only \(r<1<R\). The omitted condition \(r>0\) is necessary. If \(r\leq0\), choose \(\varepsilon>0\) small enough that \[ \varepsilon^{2/5}<R^7-R^2. \] Then the outer-circle argument in step <1>2 places all seven roots inside \(\abs{z}<R\varepsilon^{-1/5}\). Since \(p(0)=1\), every root also satisfies \(r\varepsilon^{-1/5}<\abs{z}\). Thus the printed hypotheses would put all seven roots, rather than five, in the displayed region.
Solution
Put \[ \delta \coloneqq \min\{r^2-r^7,\ R^7-R^2\}. \] Since \(0<r<1<R\), one has \(\delta>0\). Choose \(\varepsilon_0>0\) so that \[ \varepsilon_0^{2/5}<\delta, \] and fix \(0<\varepsilon<\varepsilon_0\).
1. The polynomial \(p\) has exactly two zeros, counted with multiplicity,in\[ \abs{z}<r\varepsilon^{-1/5}. \]
On the circle \[ \abs{z}=r\varepsilon^{-1/5}, \] one has \[ \begin{aligned} \abs{\varepsilon z^7+1} &\leq \varepsilon\abs{z}^7+1\\ &= r^7\varepsilon^{-2/5}+1\\ &< r^7\varepsilon^{-2/5} +(r^2-r^7)\varepsilon^{-2/5}\\ &= r^2\varepsilon^{-2/5} = \abs{z^2}. \end{aligned} \] The strict inequality uses \[ \varepsilon^{2/5}<\delta\leq r^2-r^7. \] Thus, by Rouché’s theorem, the functions \[ p(z)=z^2+(\varepsilon z^7+1) \] and \(z^2\) have the same number of zeros inside this circle. Hence \(p\) has exactly two such zeros.
2. The polynomial \(p\) has exactly seven zeros, counted with multiplicity,in\[ \abs{z}<R\varepsilon^{-1/5}. \]
On the circle \[ \abs{z}=R\varepsilon^{-1/5}, \] one has \[ \begin{aligned} \abs{z^2+1} &\leq \abs{z}^2+1\\ &= R^2\varepsilon^{-2/5}+1\\ &< R^2\varepsilon^{-2/5} +(R^7-R^2)\varepsilon^{-2/5}\\ &= R^7\varepsilon^{-2/5} = \abs{\varepsilon z^7}. \end{aligned} \] Here \[ \varepsilon^{2/5}<\delta\leq R^7-R^2. \] Rouché’s theorem therefore shows that \[ p(z)=\varepsilon z^7+(z^2+1) \] and \(\varepsilon z^7\) have the same number of zeros inside this circle, namely seven.
3. The annulus\[ r\varepsilon^{-1/5} < \abs{z} < R\varepsilon^{-1/5} \]contains exactly\[ \boxed{5} \]zeros of \(p\), counted with multiplicity.
The strict inequalities in steps 1 and 2 show in particular that \(p\) has no zero on either boundary circle. By step 2 there are seven zeros inside the outer circle, while by step 1 exactly two lie inside the inner circle. Therefore the number in the annulus is \[ 7-2=5. \]
4. Q.E.D.
Step 3 proves the asserted root count for every \(0<\varepsilon<\varepsilon_0\).