Image under a Mobius Transformation

Question

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!

Answer 1

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.

Answer 2

You already know that

• Möbius transformations map circles and lines to circles and lines, and
• Möbius transformations are uniquely identified by their values at three distinct points.

That is sufficient to solve your task. By choosing “suitable” points one can try to reduce the amount of calculation. Determining the image (or preimage) of $0$ or $\infty$ is often a good choice. I would also try to compute with complex numbers directly, and not with their cartesian coordinates.

• In order to determine the image of the real axis, we can compute $f(0) = 0$, $f(\infty) = 1+i$, and $$ f(1) = \frac{1+i}{1-i} = \frac{(1+i)^2}{(1+i)(1-i)} = \frac{2i}{2} = i \, . $$ So the image is a circle through the points $0, i, 1+i$, that is the circle with center $(1+i)/2$ and radius $1/\sqrt 2$.

  (You additionally computed $f(2) = (2+6i)/5$ and $f(3) = (3+6i)/5$. That would not have been necessary, but these points do lie on the same circle.)

• For the image of the imaginary axis we can use $f(0) = 0$, $f(\infty) = 1+i$, and $f(i) = \infty$, so the image is the line through the origin and the point $1+i$.

  (You additionally computed $f(2i) = 2+2i$ and $f(3i) = 3/2 + 3/2i$, these points lie on the same line.)

• Finally, for the image of the unit circle, we have $f(1) = i$, $f(-1) = 1$, and $f(i) = \infty$, so image is the line though the points $1$ and $i$.