I am reading on Riemann surfaces and I am trying to understand the notion of branch cut and branch points. I'm reading from Applied Complex Analysis by Yue Kuen Kwok.
Specifically, from page 121, he explains why the function $f(z)=(z^2-1)^{\frac{1}{2}}$ has two branch points $z=1$ and $z=-1$.
My understanding, vaguely speaking is that, if we pick an arbitrary point in $\mathbb{C}$ and 'circle' around either of these points (-1 or 1), the value of the argument of the function changes by $2\pi$ after we transverse angle around the point . so $z=1,-1$ are branch points.
But if we circle around both points together along a sufficiently large circle, the arguments cancel out and we end up with the original function we started with (same argument). so $z=\infty$ is not a branch point.
Also the function $f(z)=z+(z^2-1)^{\frac{1}{2}}$ has branch points $z=1,-1,\infty$ and the explanation is that the argument increases by $2\pi$ after circling around the two point $z=1$ and $z=-1$ together along a sufficiently large circle. So $z=\infty $ is one of the branch points in an addition to $z=1, z= -1$.
My question is that, the function $f(z)=z+(z^2-1)^{\frac{1}{2}}$ is one of the branches of the inverse Joukowski map, so why is $z=\infty$ NOT a branch point of the inverse Joukowski map?
Note that we have
$$\text{Res}\left(\sqrt{z^2-1},z=\infty\right)=-\text{Res}\left(\frac1{z^2}\sqrt{\frac1{z^2}-1},z=0\right)=0$$
Since the Laurent expansion of $\frac1{z^2}\sqrt{\frac1{z^2}-1}$, for $|z|<1$, has only even powers of $z^{-1}$.