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How many complex number $z$ such that $|z|< \frac{1}{3}$ and $\sum _{r=1}^{n} a_r z^r=1$, where $|a_r|<2$

Given answer is $0$ .

Could someone give me slight hint as how to approach this question?

  • 3
    **Hint:** For such a $z$ we have $\left|\sum_{r=1}^na_rz^r\right| \leq \sum_{r=1}^n 2\left(\frac 13\right)^r < 2\sum_{r=1}^\infty \left(\frac 13\right)^r$ – 2017-01-15

1 Answers 1

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If you take the asbolute value of the sum, you will get $$ |\sum_{r=1}^n a_r z^r| \leq \sum_{r=1}^n |a_r||z^r|$$ Based on the information $|a_r| <2$, you can conclude $$ \sum_{r=1}^n |a_r||z^r| < 2\sum_{r=1}^n |z|^r $$ And since $|z|<\frac{1}{3}$, you can get to $$ 2\sum_{r=1}^n |z|^r < 2\sum_{r=1}^n \frac{1}{3^r}. $$ By using the geometric series, you will arrive at $2 (\frac{1}{2} - \frac{3^{-n}}{2}) = 1 - 3^{-n}<1$. So we have, that, even if we go all the way to the most extreme values, we are allowed, we show, that abslute value of the sum is lower than one.

So there can not be equality, as long there are constraint lifted (like allowing larger $z$ or $a_r$)