# Why is $z=\infty$ not a branch point of the Inverse Joukowski map

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 $z^{-1}$.

• Please could you explain a little more? Apr 12 '18 at 3:06
• @J.Kyei Under the transformation $z\mapsto 1/z$, the residue at $z=\infty$ transforms to the residue at $z=0$. Alongside this, the function $\frac1{z^2}\sqrt{\frac1{z^2}-1}$ is an even function of $z$. Its Laurent series, therefore, cannot have any non-zero terms of odd power in $z$. In particular, the coefficient on the $z^{-1}$ term, which is the residue, must be zero. Apr 12 '18 at 3:19
• Sorry I have been away fro a while. thank you Apr 28 '18 at 3:48
• You're welcome. My pleasure. Apr 28 '18 at 12:58

It seems there is a misunderstanding. You cannot make the function $f(z) = z$ take two different values by the analytic continuation along a closed loop. Since you already know that $\sqrt {z^2 - 1}$ is unbranched over $\infty$, $z + \sqrt {z^2 - 1}$ is unbranched over $\infty$ as well.

Example 3.6.2 in Kwok's book deals with the function $-i \log(z + \sqrt {z^2 - 1})$, maybe you have forgotten the logarithm?