The number of real roots of the equation $1+a_1x+a_2x^2+.....+a_nx^n=0$ where $|x|<\frac{1}{3}$ and $|a_i|<2 \forall i=1,2,3....,n$ is The number of real roots of the equation $1+a_1x+a_2x^2+.....+a_nx^n=0$ where $|x|<\frac{1}{3}$ and $|a_i|<2  \forall i=1,2,3....,n$ is
$(a)n$ if $n$ is even
$(b)1$ if $n$ is odd. 
$(c)0$ for any natural number $n$
$(d)$none of these

As the degree of the polynomial is $n$,so it should have $n$ real roots.
But the answer in the book says number of real roots are $0$ for any natural number $n$.I dont understand why?May be $|x|<\frac{1}{3}$ and $|a_i|<2  \forall i=1,2,3....,n$ affects the number of real roots.But i dont know how?May be some Cauchy's theorem is applicable here.But i am not sure which theorem.Please help me.Thanks.
 A: Clarifying something the original poster said: When the degree of the polynomial is $n$, there are $n$ complex roots (counting multiplicity). Hence there are at most $n$ real roots.
Motivation: Noting how the range of $x$ and $a_i$ are restricted, we want to show that all the terms of $a_ix^i$ combined cannot "overpower" the $1$ in order to sum to $0$. We want a bound to prove that $1+a_1x+a_2x^2+.....+a_nx^n>0$, or equivalently, $a_1x+a_2x^2+.....+a_nx^n>-1$.
By triangle inequality,
$$|1+a_1x+a_2x^2+\dots+a_nx^n|+|a_1x|+|a_2x^2|+\dots+|a_nx^n|\\
\geq1+a_1x+a_2x^2+\dots+a_nx^n+(-a_1x)+(-a_2x^2)+(-a_3x^3)+\dots+(-a_nx^n)\\
=1$$
Let's try to substitute $x$ into the equation.
$$\begin{align*}
|1+a_1x+a_2x^2+\dots+a_nx^n|&\geq1-|a_1x|-|a_2x^2|-\dots-|a_nx^n|\\
&=1-|a_1||x|-|a_2||x|^2-\dots-|a_n||x|^n\\
&>1-2\times\frac{1}{3}-2\times\frac{1}{3^2}-2\times\frac{1}{3^3}-\dots-2\times\frac{1}{3^n}\\
&=\frac{1}{3^n}>0
\end{align*}$$
Hence, there are no real roots with magnitude $<\frac{1}{3}$.
The last equality is true as:
$$\begin{align*}
1&=3\times\frac{1}{3}\\
&=2\times\frac{1}{3}+3\times\frac{1}{3^2}\\
&=2\times\frac{1}{3}+2\times\frac{1}{3^2}+3\times\frac{1}{3^3}\\
&=\dots\\
&=2\times\frac{1}{3}+2\times\frac{1}{3^2}+2\times\frac{1}{3^3}+\dots+3\times\frac{1}{3^n}
\end{align*}$$
Rearranging gives the required expression.
