A) \[|z-a|=a\]
B) \[\left| z-\frac{1}{1-a} \right|=|1-a|\]
C) \[\left| z-\frac{1}{1-a} \right|=\frac{1}{|1-a|}\]
D) \[|z-(1-a)|\,=|\,1-a|\]
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 \[\left| {{z}_{k}}-\frac{1}{1-a} \right|=\frac{|{{a}^{k}}|}{|1-a|}=\frac{|a{{|}^{k}}}{|1-a|}<\frac{1}{|1-a|}\] Þ \[{{z}_{k}}\]lies within \[\left| z-\frac{1}{1-a} \right|=\frac{1}{|1-a|}\].You need to login to perform this action.
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