The polynomial $a z^n+z+1$ has at least one root in $|z| \leq 2$

Question

I am trying to solve this problem, but I don't have any idea.

The problem is:

Prove that for arbitrary $a \in \mathbb{C}$ and $n \geqslant 2$, polynomial $P(z) = a z^n+z+1$ has at least one root inside the disc $S = \{ z \in \mathbb{C} \mid |z| \leqslant 2 \}$.

Does anyone have idea? Thanks in advance.

Answer 1

The result follows immediately if $a=0$, so suppose $a \neq 0$.

The product of the zeros of the polynomial

$$ P(z) = az^n + z + 1 = a \left(z^n + \frac{1}{a} z + \frac{1}{a} \right) $$

is $(-1)^n/a$. If all $n$ zeros of $P$ satisfy $|z| > 2$ then this product must be greater than $2^n$ in absolute value. In other words, this would imply that

$$ \frac{1}{|a|} > 2^n, $$

which is equivalent to $|a| < 2^{-n}$. Consequently,

If $|a| \geq 2^{-n}$ then $P$ must have a zero in the disk $|z| \leq 2$.

This is part of what was asked. We need to use a different argument to handle the values of $a$ satisfying $|a| < 2^{-n}$.

Suppose that $|a| < 2^{-n}$. Then on the circle $|z| = 2$ we have

$$ |az^n| = |a|2^n < 1 $$

and

$$ |z+1| \geq |z| - |1| = 1, $$

so that $|az^n| < |z+1|$. By Rouché's theorem we conclude that the polynomial

$$ az^n+z+1 = P(z) $$

has as many zeros inside $|z| = 2$ as does the polynomial $z+1$, which is one. Restated,

If $|a| < 2^{-n}$ then $P$ has exactly one zero in the disk $|z| < 2$.

Combining these conclusions yields the desired result, that

For arbitrary $a \in \mathbb C$ and $n \in \mathbb N$ with $n \geq 2$, the polynomial $P(z) = az^n + z + 1$ has at least one zero in the disk $|z| \leq 2$.

It's worth noting that if $a=(-2)^{-n}$ then $P$ has a zero at $z=-2$ and no zeros inside $|z| < 2$, so this bound is sharp.

Answer 2

Hint: Have a look at https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem and think about what you could use for f and g (this will depend on $\alpha$). Try choosing f,g so that f + g is the equation want.

Answer 3

Take f(z)=1 and g(z)=az^n+z such that |f(z)|<|g(z)| inside S. Hence by Rouche's theorem f(z)+g(z) and g(z) has same number of zeroes (including multiplicity) inside S. As g(z) has a zero at z=0 which is always inside S, hence the result.