The polynomial $a z^n+z+1$ has at least one root in $|z| \leq 2$ 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.
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 $|z| &gt; 2$ then this product must be greater than $2^n$ in absolute value.  In other words, this would imply that

$$
\frac{1}{|a|} &gt; 2^n,
$$

which is equivalent to $|a| &lt; 2^{-n}$. 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 $|a| &lt; 2^{-n}$.

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

$$
|az^n| = |a|2^n &lt; 1
$$

and

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

so that $|az^n| &lt; |z+1|$. 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 $|a| &lt; 2^{-n}$ then $P$ has exactly one zero in the disk $|z| &lt; 2$.


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 $|z| &lt; 2$, so this bound is sharp.
Take f(z)=1 and g(z)=az^n+z such that |f(z)|&lt;|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.