Let a be a complex number such that |a| < 1 and z1, z2, Z3, ... be the vertices of a polygon such that zk = 1 + a + a2 + ... + ak- 1 for all k = 1, 2, 3, .... Then z1, z2, ... lie within the circle

Given,zk=1+a+a2+⋯+ak−1=1−ak1−a⇒zk−11−a=−ak1−a⇒zk−11−a=|a|k|1−a|<1|1−a| [∵|a|<1]Hence, zk lies within 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 of a polygon such that zk = 1 + a + a2 + ... + ak- 1 for all k = 1, 2, 3, .... Then z1, z2, ... lie within the circle

Given,zk=1+a+a2+⋯+ak−1=1−ak1−a⇒zk−11−a=−ak1−a⇒zk−11−a=|a|k|1−a|<1|1−a| [∵|a|<1]Hence, zk lies within 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 a polygon such that z_k = 1 + a + a² + ... + a^{k- 1} for all k = 1, 2, 3, .... Then z₁, z₂, ... lie within the circle

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

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

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

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

answer is A.

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Detailed Solution

Given,

$\begin{array}{l} z_{k} = 1 + a + a^{2} + \dots + a^{k - 1} = \frac{1 - a^{k}}{1 - a} \\ \Rightarrow z_{k} - \frac{1}{1 - a} = - \frac{a^{k}}{1 - a} \\ \Rightarrow |z_{k} - \frac{1}{1 - a}| = \frac{| a |^{k}}{| 1 - a |} < \frac{1}{| 1 - a |} [∵ | a | < 1] \end{array}$