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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Geometry of complex numbers

  • question_answer
    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

    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 \[\left| z-\frac{1}{1-a} \right|=\frac{1}{|1-a|}\].


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