Let \[ p(z)=2z^5+4z^2+1. \]
- How many roots does \(p\) have in the disk \(|z|<1\)?
- How many roots does \(p\) have on the real axis?
Solution
1. On the unit circle,\[ \abs{2z^5+1}<\abs{4z^2}. \]
If \(\abs z=1\), then \[ \abs{2z^5+1} \le2\abs z^5+1 =3 <4 =\abs{4z^2}. \]
2. The polynomial \(p\) has exactly\[ \boxed{2} \]zeros in \(\abs z<1\), counted with multiplicity.
By step 1 and Rouché’s theorem, \(p(z)=4z^2+(2z^5+1)\) and \(4z^2\) have the same number of zeros in the unit disk. The polynomial \(4z^2\) has exactly two zeros there, counted with multiplicity.
3. Put\[ a\coloneqq\left(\frac45\right)^{1/3}. \]The real critical points of \(p\) are \(-a\) and \(0\).
For real \(x\), \[ p'(x)=10x^4+8x=2x(5x^3+4). \] Thus \(p'(x)=0\) exactly when \(x=0\) or \(x^3=-4/5\), namely \(x=-a\).
4. The function \(p\) is strictly increasing on \((-\infty,-a)\), strictly decreasing on \((-a,0)\), and strictly increasing on \((0,\infty)\).
For \(x<-a\), both factors \(x\) and \(5x^3+4\) are negative, so \(p'(x)>0\). For \(-a<x<0\), the first factor is negative and the second positive, so \(p'(x)<0\). For \(x>0\), both are positive, so \(p'(x)>0\).
5. Both critical values are positive:\[ p(-a)>0, \qquad p(0)>0. \]
Clearly \(p(0)=1\). Since \(a^3=4/5\), \[ \begin{aligned} p(-a) &=-2a^5+4a^2+1\\ &=-2a^2a^3+4a^2+1\\ &=-\frac85a^2+4a^2+1\\ &=\frac{12}{5}a^2+1>0. \end{aligned} \]
6. The polynomial \(p\) has exactly\[ \boxed{1} \]real zero.
Since the leading term is \(2x^5\), \[ p(x)\longrightarrow-\infty \qquad\text{as }x\to-\infty. \] By step 4, \(p\) is strictly increasing on \((-\infty,-a)\), and by step 5 it ends that interval at the positive value \(p(-a)\). The intermediate value theorem therefore gives exactly one zero on \((-\infty,-a)\).
On \([-a,0]\), step 4 shows that \(p\) decreases from the positive value \(p(-a)\) to the positive value \(p(0)\), so it has no zero there. On \((0,\infty)\) it increases from \(p(0)>0\), so it has no zero there either.
7. Q.E.D.
Step 2 answers part 1, and step 6 answers part 2.