I'm currently studying Aerodynamics. A theorem very usefull that I'm learning is the Kutta-Joukowski theorem for forces and moment applied on an airfoil. I have a doubt about a mathematical step from the derivation of this theorem, which I found on a theoretical book. This step is shown on the image bellow:
(This is from a portuguese book. It's "Aerodinâmica Incompressível: Fundamentos" by Vasco Brederode http://istpress.tecnico.ulisboa.pt/node/392)
In the first part of this image there's an approximation by Taylor series expansion for the function $z=F(\zeta)$, so that we get:
$$z=F(\zeta)=\zeta+\frac{b^2}{\zeta}\hspace{15pt} (1)$$
[Note: $b$ is a real constant]
This part I totally understand. My doubt is about the final part. The inversion of the function $F(\zeta)$ it's a little obscure. I think that the author of this book did the following:
From (1) we get:
$$\zeta^2-z\zeta+b^2=0\Leftrightarrow \zeta=\frac{1}{2}\left(z\pm\sqrt{z^2-4b^2}\right)$$ where $\sqrt{z^2-4b^2}$ is the principal square root of $z^2-4b^2$.
Now, we can write $\sqrt{z^2-4b^2}=z\sqrt{1-4\left(\frac{b}{z}\right)^2}$, and expand the square root in Taylors series of $\frac{b}{z}$ around $\frac{b}{z}=0$ to give:
$$\zeta=\frac{1}{2}z\left[1\pm\sqrt{1-4\left(\frac{b}{z}\right)^2}\right]=\frac{1}{2}z\left\{ 1\pm\left[1-2\left(\frac{b}{z}\right)^2+...\right]\right\} $$
Now I think he choose the "+" branch (I wouldn't know why) and got the final result:
$$\zeta=z\left[1-\left(\frac{b}{z}\right)^2+...\right]$$
Anyway, I think that this isn't right because of the choice of the "+" branch. I reall don't find have any reason for that.
How do you interpret this step?
Well, it's a Laurent expansion, and the condition $|s/\zeta|<1$ shows we're in a neighborhood of infinity, here. We're given $$z=\zeta\left(1+\frac{b^2}{\zeta^2}+O(|\zeta|^{-3})\right),$$ this implies$$\zeta=z\left(1-\frac{b^2}{\zeta^2}+O(|\zeta|^{-3})\right),$$ since $$\frac1{1+\frac{b^2}{\zeta^2}+O(|\zeta|^{-3})}=1-\frac{b^2}{\zeta^2}+O(|\zeta|^{-3}),$$ and substituting this into itself gives$$\zeta=z\left(1-\frac{b^2}{z^2}+O(|\zeta|^{-3})\right)=z\left(1-\frac{b^2}{z^2}+O(|z|^{-3})\right),$$ since $|z/\zeta|$ will be bounded from above and below for $|\zeta|$ big enough ($\lim_{\zeta\to\infty}z/\zeta=1$).