Complex Numbers And Quadratic Equations question 322
Question: Let $ a $ be a complex number such that $ |a|<1 $ and $ z_1,z_2,…… $ 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
Options:
A) $ |z-a|=a $
B) $ | z-\frac{1}{1-a} |=|1-a| $
C) $ | z-\frac{1}{1-a} |=\frac{1}{|1-a|} $
D) $ |z-(1-a)|=|1-a| $
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Answer:
Correct Answer: C
Solution:
We have $ z_{k}=1+a+a^{2}+…..+{a^{k-1}}=\frac{1-a^{k}}{1-a} $
Þ $ z_{k}-\frac{1}{1-a}=\frac{-a^{k}}{1-a} $
Þ $ | z_{k}-\frac{1}{1-a} |=\frac{|a^{k}|}{|1-a|}=\frac{|a{{|}^{k}}}{|1-a|}<\frac{1}{|1-a|} $
Þ $ z_{k} $ lies within $ | z-\frac{1}{1-a} |=\frac{1}{|1-a|} $ .
BETA
