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Prove that for any $a\in \mathbb{C}$ and $n\geq 2$, the polynomial $az^n+z+1$ has at least one root in the disk $|z| < 2$.

My instinct is to use Rouche's Theorem for this problem, however I have only been able to prove the case when $|a| > 3/2^n$. In this case we let $f = az^n$ and $g = az^n+z+1$, and Rouche's Theorem works great. When we have $|a| \leq 3/2^n$, we run into problems. No matter what we choose for $f$ we haven't been able to get Rouche's Theorem to help us. Is this the right approach, or is there a theorem I'm forgetting that would be more useful here?

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Can you help me by any chance with my problem please? – Carpediem Feb 12 '13 at 0:54

The claim is in general not correct for $n=2$. Consider for example $a= \frac{1}{4}$:

$$a \cdot z^n +z+1 = \frac{1}{4} z^2 + z+1 = \frac{(z+2)^2}{4}$$

... and this polynomial has clearly no roots in the disk $|z|<2$.

I found proofs (using Roche's theorem) for some more special cases:

  • $|a| < \frac{1}{2^n}$: In this case let $f(z) := z$, then $$|f(z)-g(z)| = |a \cdot z^n+1| \leq a^\cdot 2^n+1 < 1+1 = 2 = |f(z)| \leq |f(z)|+|g(z)|$$
  • $|a| < \frac{3}{2^n}$, $a \in \mathbb{R}$: Let $c:= \frac{1}{2} \left(a + \frac{3}{2^n} \right)$ and $f(z) :=c \cdot z$, then $$|f(z)-g(z)| = |(a-c) \cdot z^n+z+1| \leq |a-c| \cdot 2^n+2+1 = \bigg| \underbrace{\frac{1}{2} a - \frac{1}{2} \cdot \frac{3}{2^n}}_{>0} \bigg| \cdot 2^n+3 \\ = c \cdot 2^n = |f(z)| \leq |f(z)|+|g(z)|$$
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We just discussed this problem and figured out a solution with Elise and other fellow grad students yesterday, so I'll write it here too.

As saz noted, the claim as stated is not correct. It should say "at least one root in the closed disk $|z| \leq 2$".

And again as noted in saz's answer, we need only consider the case $|a| \geq \frac{1}{2^n}$. If the roots of the polynomial are $z_1, \dots z_n$, we have $$|z_1| \dots |z_n| = \frac{1}{|a|} \leq 2^n \, .$$ Therefore at least one of the $z_i$ should satisfy $|z_i| \leq 2$.

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