How to define an analytic branch of the function $\sqrt{z^2-1}$ on the domain $\mathbb{C}\setminus [-1,1]$? We add an additional cut $[1,\infty)$ and define $\sqrt{z-1}$ on $\mathbb{C}\setminus [1,\infty)$ and $\sqrt{z+1}$ on $\mathbb{C}\setminus [-1,\infty)$ by usual way (using the formula $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\theta/2}$ for $0<\theta<2\pi$.) Now we multiply these square roots and define $\sqrt{z^2-1}=\sqrt{z-1}\times \sqrt{z+1}$ on $\mathbb{C}\setminus [-1,\infty)$. We see that the limit values of so defined $\sqrt{z^2-1}$ in the points $t+0i$ and $t-0i$, $t>1$, are equal, so we may define the function on $(1,\infty)$ by continuity and remove this camel-cut.