The Möbius transformations $f\left( z \right) = \frac{{\frac{1}{2}\left( {{k_u} - {k_l}} \right)z}}{{\frac{1}{2}\left( {{k_u} + {k_l}} \right)z + 1}}$ bijectively maps the circle that is symmetric about the real axis and cuts it (real axis) through the points $-1/k_u$ and $-1/k_l$ onto the unit circle. We also have that the real numbers $k_u, k_l$ satisfy $k_u>k_l\ge 0$.
I am trying to understand how it was derived.
Attempts: I thought about considering the original circle which is centered at $z_0=\frac{-1}{k_u}-R$ with radius $R=\frac{k_u-k_l}{2k_uk_l}$ and performing translation and then scaling it, which gives the function $f(z)=\frac{z-z_0}{R}$ but this does not give the above formula.
Any tips are appreciated.
The inversion $\,f_1(z) = \frac{1}{z}\,$ transforms the original circle into another circle $\,\Gamma\,$ of radius $\,\rho = \frac{1}{2}(k_u - k_l)\,$ and center $\,\omega=-\frac{1}{2}(k_u+k_l)\,$.
The transformation $\,f_2(z)= \frac{1}{\rho}(z-\omega)\,$ transforms $\,\Gamma\,$ to the unit circle.
The inversion $\,f_3(z)=\frac{1}{z}\,$ transforms the unit circle into itself.
It is easy to verify that $\,\frac{1}{f(z)}= \frac{1}{\rho}\left(\frac{1}{z}-\omega\right)\,$, so $\,f(z)=(f_3 \circ f_2 \circ f_1)(z)\,$ is the composition of the three transformations above, and therefore maps the original circle to the unit circle.