# Let $p(z) = az^n + z + 1$, $n \ge 2$ and $a \in \Bbb{C}$. Show that $p(z)$ has a root in the disc $|z| \le 2$.

Let $p(z) = az^n + z + 1$, $n \ge 2$ and $a \in \Bbb{C}$. Show that $p(z)$ has a root in the disc $|z| \le 2$.

I have thrown the book at this problem and have no idea how to proceed. The general plan of attack is to apply the argument principle and evaluate $\int_{|z| = 2} \frac{p'(z)}{p(z)} dz$, but since we can't assume anything about any poles located inside the contour it seems difficult to apply usual complex analytic methods, and the integral appears impossible to directly calculate. It is easy to see that if $|a| > 2^{-n}$ then the statement holds, since 1 is equal to $a$ times the product of all roots (up to multiplication by -1).

• Your proposed integral will be zero if there is no root, right? And nonzero if there is. – abnry Oct 23 '16 at 13:28
• Another track: have you seen Rouché's theorem ? See for example (math.stackexchange.com/q/1769515). – Jean Marie Oct 23 '16 at 13:32
• I have not seen Rouche's theorem, so a solution without it would be preferable. I know that it is sufficient to show the integral does not vanish, but even this is so far beyond me. – Vik78 Oct 23 '16 at 13:42
• Since $a$ is arbitrary, could induction on $n$ be useful somehow? I mean, the inductive step will be $az^{n+1} + z+1 = bz^n +z +1$ for $b=az$. So by hypotesis, this polynomial has a root in the disc. – FormerMath Oct 23 '16 at 13:48
• @abnry I understand that. As I said, I have had no luck even showing it is nonzero. – Vik78 Oct 23 '16 at 14:12

## 1 Answer

You may apply Rouché for $|a|<2^{-n}$ (since $|az^n|<1\leq |1+z|$ for $|z|=2$) and your argument with the root product for $|a|>2^{-n}$ to show that in both cases there is at least one root in the open disk of radius 2. By continuity there is at least one on the closed disk when $|a|=2^{-n}$. Note that $1+z+z^2/4=0$ shows that the radius 2 is optimal.

• Very nice-- I just needed to put the pieces together. Thanks a lot. – Vik78 Oct 23 '16 at 14:32
• Can you extend the answer by showing how to apply Rouché for $|a|<2^{-n}$? – user261263 Oct 23 '16 at 14:35
• Actually using Rouché is enough. Suppose all the roots are outside the $|z| \le 2$ disk. Then, using Vieta, one gets $|a|<2^{-n}$, impossible because of Rouché theorem – user261263 Oct 23 '16 at 14:49
• Well, I think that was the point in the root product argument (except we didn't put a name on it). – H. H. Rugh Oct 23 '16 at 15:07
• That being said, I imagine there's another way to solve this since it was a homework problem and we haven't seen Rouche's theorem yet. – Vik78 Oct 23 '16 at 19:11