Question

Let \[a\] be a complex number such that $\left| a \right|<1$ and ${{z}_{1}},{{z}_{2}},{{z}_{3}},...$ be vertices of a polygon such that ${{z}_{k}}=1+a+{{a}^{2}}+...+{{a}^{k+1}}$. Then the vertices of the polygon lie within a circle
A. \[\left| z-a \right|=a\]
B. \[\left| z-\dfrac{1}{1-a} \right|=\left| 1-a \right|\]
C. \[\left| z-\dfrac{1}{1-a} \right|=\dfrac{1}{\left| 1-a \right|}\]
D. \[\left| z-(1-a) \right|=\left| 1-a \right|\]

Answer 1

Hint: In this question, we have to find the condition that represents the vertices of a polygon within the circle. Since the given series is a geometric series, we can able to extract the given condition by using the sum of the terms in the series.

Formula Used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
If a series is in geometric progression, then its sum is
$\begin{align}
  & {{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r} \\
 & r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}} \\
\end{align}$
Here $r$ is the common ratio of the terms in the series.

Complete step by step solution: Given that \[a\] is a complex number. So, $\left| a \right|<1$
The given series for the vertices of a polygon is
${{z}_{k}}=1+a+{{a}^{2}}+...+{{a}^{k+1}}$
Since the given series is a geometric series, we can able to find its the sum by the formula
$\begin{align}
  & {{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r} \\
 & r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}} \\
\end{align}$
Here the common ratio is
$r=\dfrac{{{a}^{2}}}{a}=a$
Thus,
${{S}_{k}}=\dfrac{1(1-{{a}^{k}})}{1-a}$
So, we can write
$\begin{align}
  & {{z}_{k}}=1+a+{{a}^{2}}+...+{{a}^{k+1}}=\dfrac{1(1-{{a}^{k}})}{1-a} \\
 & \Rightarrow {{z}_{k}}=\dfrac{(1-{{a}^{k}})}{1-a} \\
 & \Rightarrow {{z}_{k}}=\dfrac{1}{1-a}-\dfrac{{{a}^{k}}}{1-a} \\
 & \Rightarrow {{z}_{k}}-\dfrac{1}{1-a}=\dfrac{-{{a}^{k}}}{1-a} \\
\end{align}$
Since their magnitude becomes,
$\left| {{z}_{k}}-\dfrac{1}{1-a} \right|=\dfrac{\left| {{a}^{k}} \right|}{\left| 1-a \right|}$
Since \[{{a}^{k}}\] is a complex number, \[\left| {{a}^{k}} \right|<1\]
So,
$\dfrac{\left| {{a}^{k}} \right|}{\left| 1-a \right|}<\dfrac{1}{\left| 1-a \right|}$
Since the given vertices ${{z}_{1}},{{z}_{2}},{{z}_{3}},...{{z}_{k}}$ lie within the circle, we can write
$\left| {{z}_{k}}-\dfrac{1}{1-a} \right|=\dfrac{1}{\left| 1-a \right|}$

Option ‘C’ is correct

Note: Here we need to remember that the given vertices of the polygon lie within the circle and the given series is a geometric series with complex numbers. So, by using the sum of the geometric series, we can able to extract the required condition.