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 $\mathcal{Re}(z)=0$ ?

In triangle inequality $|z_1+z_2|\leq |z_1|+|z_2|$ we have equality iff $z_1=0$ or there is a $\alpha\geq 0$ with $z_2=\alpha z_1$. Then, $$|z^2-1|=|z|^2+1\Leftrightarrow |z^2+(-1)|=|z^2|+|-1|$$ $$\Leftrightarrow z^2=\alpha (-1)\le 0\Leftrightarrow z=\pm\sqrt{\alpha}i$$ so, $z$ belongs to the imaginary axis.
