Question

Let $a$ be complex number such that $\left|a\right|<1$ and $z_1,z_2,z_3,...$ be the vertices of a polygon such that $z_k=1+a+a^2+..+a^{k-1}$ for all $k=1,2.3,...$ Then $z_1,z_2,...$ lie within the circle

• A. $|z-\frac{1}{1-a}|=\frac{1}{|a-1|}$
• B. $|z+\frac{1}{1+a}|=\frac{1}{|a+11|}$
• C. $|z-\frac{1}{1-a}|=|a-1|$
• D. $|z+\frac{1}{1+a}|=|a+1|$

Answer 1

The correct option is A $|z-\frac{1}{1-a}|=\frac{1}{|a-1|}$

Here $z_k$ forms a G.P.
Simplifying, we get
$z_k=\frac{1-a^k}{1-a}$
Thus
$z_k-\frac{1}{1-a}=\frac{-a^k}{1-a}$
Now
$|\frac{-a^k}{1-a}|<|\frac{1}{1-a}|$
Hence $z_1,z_2..$ lie within the circle :-
$|z_k-\frac{1}{1-a}|=|\frac{1}{1-a}|$