Q. Let $a$ be a complex number such that $|a| < 1$ and $\mathcal{z}_{1},\ \mathcal{z}_{2},\ldots$ be vertices of a polygon such that $\mathcal{z}_{k} = 1 + a + a^{2} + ... + a^{k - 1}$. Then the vertices of the polygon lie within a circle is
A) $\left| \mathcal{z -}a \right| = a$
B) $\left| \mathcal{z -}\frac{1}{1 - a} \right| = |1 - a|$
C) $\left| \mathcal{z -}\frac{1}{1 - a} \right| = \frac{1}{|1 - a|}$
D) $\left| \mathcal{z -}(1 - a) \right| = |1 - a|$
Explanation
(c)
We have, $\mathcal{z}_{k} = 1 + a + a^{2} + ... + a^{k - 1} = \frac{1 - a^{k}}{1 - a}$
$\Rightarrow \ \mathcal{z}_{k} - \frac{1}{1 - a} = \frac{- a^{k}}{1 - a}$
$\Rightarrow \ \left| \mathcal{z}_{k} - \frac{1}{1 - a} \right| = \frac{|a^{k}|}{|1 - a|}$
$= \ \frac{|a^{k}|}{|1 - a|} < \frac{1}{|1 - a|}\ \ \ \ (\because|a| < 1)$
$\Rightarrow \ \mathcal{z}_{k}$ lies within a circle $\left| \mathcal{z -}\frac{1}{1 - a} \right| = \frac{1}{|1 - a|}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.