Is it true that $ \sqrt{z^2-1} = i \sqrt{1-z^2} $? I have seen a lot of times in books or on the internet that$ \sqrt{z^2-1} = i \sqrt{1-z^2} $   and I don't understand why that is correct . In general it is not true that $ \sqrt{-z}=i  \sqrt {z}$ and I think that we use that to show the above.
I think if $(1-z^2)$ has argument in $(-π,0]$, then we have that $ \sqrt{-z}=i  \sqrt {z}$ but it's not always like that, for example if $z=-2+i$ then $ z^2=4-4i-1$ so $ 1-z^2=-2+4i $ and that one has not the argument that we want. 
Thank you in advance!

The problem is the definition of the square root. Every complex number has 2 square roots (except 0), and we agree that if the number is real and positive, the sign $ \sqrt ?$ means to take the positive root. But when we're dealing with a general complex number z, we don't have one good way to define which of the roots will be $\sqrt z$. So for a general $z$, the eqation you wrote isn't well defined.
However, it is true that $\frac{\sqrt {1-z^2}}{\sqrt {z^2-1}}$ will always be $i$ or $-i$, no matter which root you choose for those numbers.
