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Prove that there exists no complex no. $z$ such that $\displaystyle |z|<\frac{1}{3}$ and $\displaystyle \sum^{n}_{r=1}a_{r}z^{r} = 1\;,$ Where $|a_{r}|<2$

$\bf{My\; Try::}$ We can write $\displaystyle \sum^{n}_{r=1}a_{r}z^{r} = 1$ as

$$\displaystyle a_{1}z+a_{2}z^2+a_{3}z^3+..............a_{n}z^{n} = 1$$

Now Using $\triangle$ Inequality, We get

$$\displaystyle |a_{1}z+a_{2}z^2+a_{3}z^3+.............+a_{n}z^n|\leq |a_{1}z|+|a_{2}z^2|+|a_{3}z^3|+...........+|a_{n}z^n|$$

Now How Can I solve after that, Help me

Thanks

juantheron
  • 53,015

2 Answers2

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Let $|a_{\text{max}}|=$max$\{a_r, 1\le r\le n\}$

$$|a_{1}z|+|a_{2}z^2|+|a_{3}z^3|+...........+|a_{n}z^n|$$

$$\le a_{\text{max}}(|z|+|z|^2+\cdots+|z|^n)=a_{\text{max}}\dfrac{|z|-|z|^{n+1}}{1-|z|}$$

$$\le a_{\text{max}}\dfrac{|z|}{1-|z|}$$

$$= a_{\text{max}}\left(\dfrac1{1-|z|}-1\right)$$

If $|z|<r<1, -|z|>-r\iff1-|z|>1-r\iff\dfrac1{1-|z|}<\dfrac1{1-r}$

$\iff\dfrac1{1-|z|}-1<\dfrac1{1-r}-1=\dfrac r{1-r}$

lab bhattacharjee
  • 274,582
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Suppose $\displaystyle \sum^{n}_{r=1}a_{r}z^{r} = 1$. Then $z\neq 0$.

Now, using the triangle inequality, the formula for the sum of a geometric series, and the condition on the $a_r,\ $we have

$1=\left | \sum^{n}_{r=1}a_{r}z^{r} \right |\leq 2\frac{\vert z\vert -\vert z\vert^{n+1}}{1-\vert z\vert}< \frac{2\vert z\vert }{1-\vert z\vert },\ $ where the last inequality is strict because $z\neq 0$. But now observe that, the condition on $z$ implies that $\frac{2\vert z\vert }{1-\vert z\vert }\leq \frac{2/3}{1-1/3}=1$

and so we conclude that $1=\left | \sum^{n}_{r=1}a_{r}z^{r} \right |<1$, which is a contradiction.

Matematleta
  • 29,139