If $|z^2-1|=|z|^2+1$, show that $z$ lies on imaginary axis If $|z^2-1|=|z|^2+1$, how do we show that $z$ lies on imaginary axis ?
I understand that I can easily do this if I substitute $z=a+ib$. How do we solve it using algebra of complex numbers without the above substitution ?
My Attempt:
$$
|z|^2+|1|^2=|z-1|^2+2\mathcal{Re}(z)=|z^2-1|\\
2\mathcal{Re}(z)=|z^2-1|-|z-1|^2=|(z+1)(z-1)|-|z-1|.|z-1|\\=|z+1|.|z-1|-|z-1|.|z-1|
$$
How do I proceed further and prove $\mathcal{Re}(z)=0$ ?

Use the fact that $|z|^2 = z\bar{z}$.
Squaring both sides of the given equality yields
\begin{align}
|z^2-1|^2 &= (z\bar{z} + 1)^2\\
(z^2 - 1)(\bar{z}^2 - 1) &= (z\bar{z} + 1)(z\bar{z}+1)\\
z^2 + 2z\bar{z} + \bar{z}^2 &= 0\\
(z + \bar{z})^2 &= 0\\
z = -\bar{z}
\end{align}
from which it follows that the real part of $z$ is $0$. (I skipped some simple algebra steps above.)
