Let z=x+iy⇒z2=x2−y2+2ixy Now, 2iz1+z2=2i(x+iy)1+x2−y2+2ixy=2ix−2y(x2−y2+1)+2ixy=−2y+2ix(x2−y2+1)+2ixy×−2y−2ix−2y−2ix=2(x2+y2)y(x2+y2−1)+x(x2+y2+1)i a=2(x2+y2)x(x2+y2+1) Since, ∣z∣=1⇒x2+y2=1 ⇒x2+y2=1∴a=2×1x(1+1)=x Also z=1⇒x+iy=1∴A=(−1,1)