Question

Let $a$ be a complex number such that $\left|a\right|<1$ and $z_1,z_2,...,z_n$ be the vertices of a polygon such that $z_k=1+a+a^2+...+a^n$, then the vertices of the polygon lie within the circle

• A. $\left|z\right|=\frac{1}{|1-a|}$
• B. $|z-a|=\frac{1}{|1-a|}$
• C. $|z-\frac{1}{1-a}|=\frac{1}{|1-a|}$
• D. None of these

Answer 1

Answer: (C)

$|z-\frac{1}{1-a}|=\frac{1}{|1-a|}$

We have, $z_k=1+a+a^2+...+a^n=\frac{1-a^{k+1}}{1-a}$

$\Rightarrow z_k-\frac{1}{1-a}=\frac{-a^{k+1}}{1-a}$

$\Rightarrow |z-\frac{1}{1-a}|=\frac{{\left|a\right|}^{k+1}}{|1-a|}<\frac{1}{|1-a|}(\because \left|a\right|<1)$

Therefore, vertices $z_1,z_2,...,z_n$ of the polygon lie within the circle.

$|z-\frac{1}{1-a}|=\frac{1}{|1-a|}$