Given that ∣z2−1∣=∣z∣2+1⇒∣z2+(−1)∣=∣z2∣+∣−1∣ It shows that the origin, −1 and z2 lies on a line and z2 and −1 lies on one side of the origin, therefore z2 is a negative number. Hence z will be purely imaginary. So we can say that z lies on y-axis. Alternate Solution We know that, if ∣z1+z2∣=∣z1∣+∣z2∣⇒arg(z1)=arg(z2)∵∣z2+(−1)∣=∣z2∣+∣−1∣⇒arg(z2)=arg(−1)⇒2arg(z)=π(∵arg(−1)=π)⇒arg(z)=2π⇒z lies on y-axis (imaginary axis).