Let z=32+i25+32−i25 .If R(z) and I(z) respectively denote the real and imaginary parts of z, then

Let z=32+i25+32i25 .If R(z) and I(z) respectively denote the real and imaginary parts of z, then

  1. A

    R(z)>0 and I(z)>0 

  2. B

    I(z)=0 

  3. C

    R(z)<0 and I(z)=0

  4. D

    R(z)=3

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    Solution:

    We have z=32+i25+32i23

    =cosπ6+isinπ65+cosπ6isinπ65=cos5π6+isin5π6+cos5π6isin5π6=2cos5π6<0

    Here, I(z)>0 and R(z)>0.

     

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