|2=(|z|2+1)2 ⇒(z2−1)(z2−1)=(|z|2+1)2[. as |z|2=zz] ⇒(z2−1)((z)2−1)=(|z|2+1)2 ⇒(zz)2−z2−(z)2+1=|z|4+2|z|2+1 ⇒|z|4−z2−(z)2+1=|z|4+2|z|2+1 ⇒z2+(z)2+2zz=0 ⇒(z+z)2=0 ⇒z+z=0 ⇒z=−z If z=x+iy then z=x-iy ‌∴x+iy=−(x−iy) ‌⇒x+iy=−x+iy ‌⇒x=0 ∴z is purely imaginary. So, it is lie on the imaginary axis.