Let z=x+iy z+iz−i is purely imaginary means its real part is zero. x+iy+ix+iy−i=x+i(y+1)x+i(y−1)×x−i(y+1)x−i(y+1)=x2+(y+1)2x2−2ix(y+1)+xi(y−1)+y2−1=x2+(y+1)2x2+y2−1−x2+(y+1)22xi for pure imaginary, we have ⇒⇒⇒x2+(y+1)2x2+y2−1=0x2+y2=1(x+iy)(x−iy)=1x+iy=x−iy1=z and z1=x−iy z+z1=(x+iy)+(x−iy)=2x (z+z1) is any non-zero real number