Let \(a\in\mathbb C\) and let \(n\ge2\) be an integer. Prove that \[ 1+z+az^n=0 \] has at least one root in the closed disk \[ |z|\le2. \]
Solution
1. If \(a=0\), the equation has the root \(z=-1\).
For \(a=0\), the equation is \(1+z=0\), and \[ \abs{-1}=1\leq2. \]
2. Suppose \(a\neq0\), set \(b\coloneqq a^{-1}\), and define\[ p(z)\coloneqq z^n+bz+b. \]Then \(p\) has exactly the same roots as \(1+z+az^n\).
Multiplying the original polynomial by the nonzero scalar \(b\) gives \[ b(1+z+az^n)=b+bz+z^n=p(z). \] Multiplication by a nonzero scalar does not change the zero set.
3. If \(\abs b\leq2^n\), then \(p\) has a root \(z_0\) with\(\abs{z_0}\leq2\).
Let \(z_1,\ldots,z_n\) be the roots of the monic polynomial \(p\), counted with multiplicity. By Vieta’s formula, \[ \abs{z_1\cdots z_n}=\abs b. \] If every root satisfied \(\abs{z_j}>2\), then \[ \abs{z_1\cdots z_n}>2^n, \] contradicting \(\abs b\leq2^n\).
4. If \(\abs b>2^n\), then \(p\) has a root \(z_0\) with\(\abs{z_0}<2\).
On the circle \(\abs z=2\), let \[ g(z)\coloneqq b(1+z). \] The reverse triangle inequality gives \[ \abs{1+z}\geq\bigl|\abs z-1\bigr|=1. \] Hence \[ \abs{p(z)-g(z)} =\abs{z^n} =2^n <\abs b \leq\abs{b(1+z)} =\abs{g(z)}. \] By Rouché’s theorem, \(p\) and \(g\) have the same number of zeros in \(\abs z<2\), counted with multiplicity. The function \(g\) has exactly one zero there, namely \(z=-1\). Thus \(p\) has a zero in \(\abs z<2\).
5. For every \(a\in\CC\) and every integer \(n\geq2\),\[ \boxed{\text{there exists }z_0\in\CC\text{ such that } 1+z_0+az_0^n=0\text{ and }\abs{z_0}\leq2}. \]
Step 1 handles \(a=0\). For \(a\neq0\), either \(\abs b\leq2^n\) or \(\abs b>2^n\); steps 3 and 4 respectively give the required root, and step 2 transfers that root to the original polynomial.
6. Q.E.D.
Step 5 is exactly the required conclusion.