Let ‘a’ be a complex number such that a<1 and Z1, Z2, Z3 ........be the vertices ofpolygon such that Zk = 1 + a + a2 + ......... + ak-1 for all k = 1,2,3…….then the verticesof the polygon lie with in a circle.

Zk=1+a+a2+….+ak−1=1−ak1−a⇒Zk−11−a=−ak1−aZk−11−a=−ak1−a=|a|k|1−a|<1|1−a|⇒ZK lies with in the Circle Z−11−a=1|1−a|

Let ‘a’ be a complex number such that a<1 and Z1, Z2, Z3 ........be the vertices ofpolygon such that Zk = 1 + a + a2 + ......... + ak-1 for all k = 1,2,3…….then the verticesof the polygon lie with in a circle.

Zk=1+a+a2+….+ak−1=1−ak1−a⇒Zk−11−a=−ak1−aZk−11−a=−ak1−a=|a|k|1−a|<1|1−a|⇒ZK lies with in the Circle Z−11−a=1|1−a|

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Let ‘ $a$ ’ be a complex number such that $|a| <$ 1 and Z₁, Z₂, Z₃ ........be the vertices of
polygon such that Z_k = 1 + a + a² + ......... + a^k-1 for all k = 1,2,3…….then the vertices
of the polygon lie with in a circle.

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$| Z - a | = | a |$

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

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

$| Z - 1 + a | = | 1 - a |$

answer is C.

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Detailed Solution

$\begin{array}{l} Z_{k} = 1 + a + a^{2} + \dots . + a^{k - 1} = \frac{1 - a^{k}}{1 - a} \\ \Rightarrow Z_{k} - \frac{1}{1 - a} = \frac{- a^{k}}{1 - a} \\ |Z_{k} - \frac{1}{1 - a}| = |\frac{- a^{k}}{1 - a}| = \frac{| a |^{k}}{| 1 - a |} < \frac{1}{| 1 - a |} \end{array}$
$\Rightarrow Z_{K}$ lies with in the Circle $|Z - \frac{1}{1 - a}| = \frac{1}{| 1 - a |}$