<s>[INST] Image under a Mobius Transformation I am studying the Mobius transformation
$$f(z) = \frac{z(1+i)}{z - i}$$
I need to find where the transformation maps the real axis, the imaginary axis and the unit circle. I know that Mobius transformations map lines and circles to lines or circles.
So I let $z = x + iy$, so we have $$f(x,y) = \frac{(x+iy)(1 + i)}{x + iy - i}$$ Now the real line is given by $y = 0$, and the imaginary axis by $x = 0$ and the unit circle by $x^2 + y^2 = 1$. From here, I substitute each of these values:
$$f(x, 0) = \frac{(x)(1+i)}{x - i}$$
Now I have seen that we only need three points to uniquely determine a line/circle so I tested several values to see if I could see any pattern
$$f(-1,0) = 1$$
$$f(0,0) = 0$$
$$f(1,0) = i$$
$$f(2,0) = \frac{2 + 6i}{5}$$
$$f(3,0) = \frac{3 + 6i}{5}$$
I graphed these values using Desmos, and this does not appear to be a line or a circle.
I did the same for $$f(0,y) = \frac{(iy)(1 + i)}{iy - i}$$ to get
$$f(0,-1) = \frac{1+i}{2}$$
$$f(0,0) = 0$$
$$f(0,1) = \infty$$
$$f(0,2) = 2+2i$$
$$f(0,3) = \frac{3 + 3i}{2}$$
Again, I do not see how this is a line or a circle.
Finally, I am not exactly sure what to do with the unit circle. Although I could plug in some values for the that lie on the unit circle, it does not seem that my approach is working.
I appreciate any help!
 [/INST] Start with
$$
\frac{z}{z-i} = \frac{x+iy}{x+(y-1)i} = \frac{(x+iy)(x-(y-1)i)}{x^2+(y-1)^2} = \\
= \frac{x^2 + y(y-1) + (xy - x(y-1))i}{x^2+(y-1)^2} = 
\frac{x^2 + y(y-1)}{x^2+(y-1)^2} + i\frac{x}{x^2+(y-1)^2}
$$
So, for $z = x + i 0$, $x\in\mathbb{R}$:
$$
f(z) = (1+i)\left(
\frac{x^2}{x^2+1} + i\frac{x}{x^2+1}
\right)
$$
Since $\frac{x^2}{x^2+1} = \frac{x^2}{x^2+1} - \frac{1}{2} + \frac{1}{2} = \frac{2x^2-x^2-1}{2x^2+2} + 
\frac{1}{2} = \frac{x^2 -1}{2x^2+2} + \frac{1}{2} = \frac{1}{2}\frac{x^2-1}{x^2+1}+\frac{1}{2}$,
$$
f(z) = (1+i)\cdot\frac{1}{2}\cdot\left(\frac{x^2-1}{x^2+1} + i \frac{2x}{x^2+1} + 
1\right)
$$
As $\left(\frac{x^2-1}{x^2+1}\right)^2 + \left(\frac{2x}{x^2+1}\right)^2 = 
\frac{x^4 - 2x^2 + 1 + 4x^2}{x^4 + 2x^2 + 1} = 1$, one may guess that
$\left(\frac{x^2-1}{x^2+1}, \frac{2x}{x^2+1}\right), x\in\mathbb{R}$ curve is unit circle centered at zero, so $\frac{x^2}{x^2+1} + i\frac{x}{x^2+1}$ gives circle with radius $\frac{1}{2}$ centered at $(1, 0)$, and multiplying it with $1+i$ simply scales it and rotates, since $1+i = \sqrt{2}e^{i\pi/4}$.
Check it in Desmos. So, image of real line is cirlce with radius $\frac{\sqrt{2}}{2}$ and center in $(0.5, 0.5)$.
Similarly, for imaginary axis with $z = 0 + iy$, $y\in\mathbb{R}$:
$$
f(z) = (1+i)\left(\frac{y(y-1)}{(y-1)^2}\right) = (1+i)\frac{y}{y-1}
$$
So image of imaginary axis is line through $0$ and $(1, 1)$ points on a plane.
Finally, case of unit cirlce.
$$
\frac{z}{z-i} = 
\frac{x^2 + y(y-1)}{x^2+(y-1)^2} + i\frac{x}{x^2+(y-1)^2} = 
\frac{x^2 + y^2 - y}{x^2 + y^2 - 2y + 1} + i \frac{x}{x^2+y^2 -2y+1}
$$
For $z = e^{it}$, $t\in [0; 2\pi]$, $x^2 + y^2 = 1$, so
$$\frac{z}{z-i} = \frac{1-y}{2 - 2y} + i\frac{x}{2-2y} = \frac{1}{2} + i\cdot\frac{1}{2}\cdot\frac{\cos t}{1 - \sin{t}}$$
$\frac{\cos t}{1 - \sin{t}}$ is clearly unbounded, so it is equation of line $
\frac{1}{2} + i y$, $y\in\mathbb{R}$. Multiplying by $1+i$, you get
$\frac{1}{2} - y + i\left(\frac{1}{2} +y\right)$ - it's also a straight line, check it with Desmos.
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