# If $\;|z^{2}-1| = |z|^{2}+1\;$, then show that $z$ lies on imaginary axis

$\begin{array}{1 1} \text{(a) z lies on imaginary axis} \\ \text{(b) z lies on real axis} \\ \text{(c) z is not lies on imaginary axis} \\ \text{(d)None of these}\end{array}$

Answer : z lies on imaginary axis
Explanation :
Let $\;z=x+iy$
Then , $\;|z^{2}-1| = |z|^{2}+1\;$
$|x^{2}-y^{2}-1+i2xy | = |x+iy|^{2}+1$
$(x^{2}-y^{2}-1)^{2}+4x^{2}y^{2} =(x^{2}+y^{2}+1)^{2}$
$4x^{2}=0\;$ i.e x=0
Hence z lies on y-axis