Question

If $\left|a_i\right|<2,i\in \{1,2,\mathrm{3...}n\}$. Prove that for no $z$, $\left|z\right|<\frac{1}{3}$ and $\sum _{i=1}^n{a}_i{z}^i=1$ can occur simultaneously.

Answer 1

Given $\left|{\alpha }_i\right|<2\forall i$ .....$\left(1\right)$
$\left|z\right|<\frac{1}{3}$ .....$\left(2\right)$
$\sum _{i=1}^n{\alpha }_i{z}^i=1$ .....$\left(3\right)$
Now we know that modulus of sum $\le$ sum of moduli and modulus of product is equal to product of moduli.
$1=\left|\sum _{i=1}^n{\alpha }_i{z}^i\right|\le \sum \left|{\alpha }_i{z}^i\right|=\sum \left|{\alpha }_i\right|\left|z^i\right|<2\sum z^i$ by $\left(1\right)$
$=2[\left|z\right|+{\left|z\right|}^2+{\left|z\right|}^3+........+{\left|z\right|}^n]$
Divided by $2$ and add $1$ in both sides
$\therefore \frac{1}{2}+1<1+\left|z\right|+{\left|z\right|}^2+.....{\left|z\right|}^n$
$\therefore 1+\left|z\right|+{\left|z\right|}^2+.....{\left|z\right|}^n>\frac{3}{2}$ or $S_n>\frac{3}{2}$
Case I. If $\left|z\right|<1$, then
$S_{\infty }>S_n\Rightarrow \frac{1}{1-\left|z\right|}>\frac{3}{2}$ or $2>3-3\left|z\right|$
$\Rightarrow \left|z\right|>\frac{1}{3}$And this violates $\left(2\right)$.
Case II. If $4\left|z\right|\ge 1$, then both $\left(2\right)$ and $\left(3\right)$ cannot hold simultaneously. Also it violates $\left(2\right)$.