Complex Numbers 2 Question 38
39.
Prove that there exists no complex number $z$ such that $|z|<1 / 3$ and $\sum _{r=1}^{n} a _r z^{r}=1$, where $\left|a _r\right|<2$.
(2003, 2M)
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Solution:
- Given, $a _1 z+a _2 z^{2}+\ldots+a _n z^{n}=1$
$ \begin{array}{lc} \text { and } & |z|<\frac{1}{3} \\ \therefore & \left|a _1 z+a _2 z^{2}+a _3 z^{3}+\ldots+a _n z^{n}\right|=1 \\ \Rightarrow & \left|a _1 z\right|+\left|a _2 z^{2}\right|+\left|a _3 z^{3}\right|+\ldots+\left|a _n z^{n}\right| \geq 1 \end{array} $
[using $\left|z _1+z _2\right| \leq\left|z _1\right|+\left|z _2\right|$ ]
$\Rightarrow \quad 2(\left|z|+|z|^{2}+|z|^{3}+\ldots+|z|^{n} )>1 \quad\right.$
[using $\left.\left|a _r\right|<2\right]$
$\Rightarrow \quad \frac{2|z|\left(1-|z|^{n}\right)}{1-|z|}>1 \quad$ [using sum of $n$ terms of GP]
$\Rightarrow \quad 2|z|-2|z|^{n+1}>1-|z|$
$ \begin{array}{ll} \Rightarrow & 3|z|>1+2|z|^{n+1} \\ \Rightarrow & |z|>\frac{1}{3}+\frac{2}{3}|z|^{n+1} \\ \Rightarrow & |z|>\frac{1}{3}, \text { which contradicts } \end{array} $
$\therefore$ There exists no complex number $z$ such that
$ |z|<1 / 3 \text { and } \sum _{r=1}^{n} a _r z^{r}=1 $
