TITLE: Find all solutions to $|z+\sqrt{z^2-1}|=1$
QUESTION [5 upvotes]: I'm trying to find the solution set to $$|z+\sqrt{z^2-1}|=1$$
I'm getting $z\in [-1,1]$, which is making me a little paranoid that it's purely a real number.  Can you check my work?
$$\begin{align}|z+\sqrt{z^2-1}|=1 &\implies z+\sqrt{z^2-1} = e^{i\theta} \\ &\implies z^2-1 = (e^{i\theta}-z)^2=e^{i2\theta}-2ze^{i\theta}+z^2 \\ &\implies 2ze^{i\theta} = e^{i2\theta}+1 \\ &\implies z = \frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right) = \cos(\theta) \\ &\implies z \in [-1,1]\end{align}$$

REPLY [2 votes]: You have successfully proved that $z \in [-1, 1]$. You should also prove that every $z$ in that range is actually a solution. This isn't that hard, but you can't do it by just running your current proof backward; you squared both sides of your equation, which might have introduced extraneous solutions.