I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = \frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be turned into inversion as well:
- $C_1= 4|z|^2+3\overline{z}-3z-2 $
- $C_2 =|z|^2 – 1$
Or we could turn the second circle into a fractional linear transforation $g(z) = – \frac{1}{z}.$
Then I could multiply the two transformations:
$$
\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right]
\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]
=
\left[ \begin{array}{cc} 2 & 3 \\ 3 & 4 \end{array} \right]
$$
and this could turn back into a circle:
- $ C_1C_2 = 3|z|^2 + 4 \overline{z} + 3z + 2 $
I found this technique in a somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and circles can be identified.
Best Answer
In the present case, $w=f(z)$ means that $$w=\frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=\frac{2-3w}{4w-3}$$ Thus, the image of the circle has equation $$\left|\frac{2-3w}{4w-3}\right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+\bar w)+9|w|^2=16|w|^2-12(w+\bar w)+9$$ that is, $$7|w|^2-6(w+\bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=\frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.
Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...