If a>0 and z=(1+i)2a−i, has magnitude 25, then z- is equal to :

If a>0 and z=(1+i)2ai, has magnitude 25, then z- is equal to :

  1. A

    1535i

  2. B

    3515i

  3. C

    1535i

  4. D

    15+35i

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    Solution:

    We have, z=(1+i)2ai

    =2iai=2i(a+i)a2+1=2a2+1+2aia2+1

    Since  |z|=2/5 [Given]

    4a2+12+4a2a2+12=2521+a2=25
    41+a2=25                                                           [Squaring both sides]

     1+a2=10a2=9a=±3a=3[a>0]z=15+35iz¯=1535i

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