Let a be a complex number such that |a|

Question

(C) We have $z_k = 1 + a + a^2 + \dots + a^{k-1} = \frac{1 - a^k}{1 - a}$.Subtracting $\frac{1}{1 - a}$ from both sides,we get:$z_k - \frac{1}{1 - a}

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

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(C) We have $z_k = 1 + a + a^2 + \dots + a^{k-1} = \frac{1 - a^k}{1 - a}$.Subtracting $\frac{1}{1 - a}$ from both sides,we get:$z_k - \frac{1}{1 - a} = \frac{1 - a^k}{1 - a} - \frac{1}{1 - a} = \frac{-a^k}{1 - a}$.Taking the modulus on both sides:$\left| z_k - \frac{1}{1 - a} \right| = \left| \frac{-a^k}{1 - a} \right| = \frac{|a|^k}{|1 - a|}$.Since $|a| Therefore,$\left| z_k - \frac{1}{1 - a} \right| This implies that the vertices $z_k$ lie within the circle defined by $\left| z - \frac{1}{1 - a} \right| = \frac{1}{|1 - a|}$.

Let $a$ be a complex number such that $|a| < 1$ and $z_1, z_2, \dots$ be vertices of a polygon such that $z_k = 1 + a + a^2 + \dots + a^{k-1}$. Then the vertices of the polygon lie within a circle:

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