Let z = 3 2 + i 2 5 + 3 2 - i 2 5 . If R z and I z respectively denote the real and imaginary parts of z ,…

Let z=32+i25+32-i25. If Rz and Iz respectively denote the real and imaginary parts of z, then
  1. Iz=0
  2. Rz<0 and Iz>0
  3. Rz>0 and Iz>0
  4. Rz=-3

Solution

Let α=32+i2, then z=α5+α¯5 

We know z¯n=z¯n , 

then, z=2Reα5 

Hence, Iz=0.

Asked in: JEE Main 2019 (10 Jan Shift 2)

Practice more Complex Number questions on Aicharya