Finding a Möbius Transformation given constraints

Question

I am trying to solve this problem, but am running into very complicated solving, and think that there is a simpler approach that I am missing.

Find a Möbius transformation $M(z)$ that satisfies the following properties:

• The image of the circle $|z|=2$ is the line $\def\re{\operatorname{Re}}\re(z) = \def\im{\operatorname{Im}}\im(z)$.
• The region where $|z|<2$ maps to the region where $\re(z) > \im(z)$.
• The point $\sqrt2+i\sqrt2 $ is a fixed point.

(original scan)

I tried selecting 3 points and constructing a unique transformation that way. I picked $\sqrt{2}+i\sqrt{2}$ because that is the fixed point, and I also picked $M(2)=\frac{3\sqrt{2}}{2}+i\frac{3\sqrt{2}}{2}$ and $M(i)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$. I was hoping that when I found it, it would map the inside of the circle to the lower half of the plane below the line $\re{z}=\im{z}$, but I hadn't gotten that far yet, because of the very complicated solving that needs to be done in order to use this method. Is there something I am missing? I feel like there is a much simpler way of solving this.

Also, I was thinking that there is not a unique Möbius transformation that satisfies these constraints, as there are (many) more than one 3-point sets you could find a transformation for that satisfies this. Thoughts?

Thanks!

Answer 1

I would try to find a function of the form $$f(z) = \rho T(z) + s(1+i)$$ where $|\rho|=1$, $s\in\mathbb R$, and $T$ is a Mobius transformation mapping the circle $|z|=2$ to a line. It shouldn't be too hard to find $T$; just include a $z-2$ in the denominator. You can then multiply by $\rho$ to rotate and shift by $s$ to get the desired fixed point.

Specifically, let's take $T(z)=(z-2)/(z+2)$. This maps $2\mapsto0$, $-2\mapsto\infty$, and $0\mapsto-1$. As result, the circle $|z|=2$ maps to the imaginary axis and it's interior maps to the left half-plane $Im(z)\leq0$. We can then multiply by $e^{3\pi i/4}$ to rotate the imaginary axis to the line $Re(z)=Im(z)$ that you want. If you stop there, you will find that $$(1+i)\sqrt{2} \mapsto -\frac{1+i}{2+\sqrt{2}},$$ so just shift back by that amount plus $(1+i)\sqrt{2}$. Putting this all together, we get $$M(z)=e^{3\pi i/4}\frac{z-2}{z+2}+\frac{3+2 \sqrt{2}}{2+\sqrt{2}}(1+i).$$ We can illustrate the three steps of inversion, rotation, and translation to see what's going on.

You are correct that this is not unique, as you can compose this function with any scaling about the fixed point to obtain another Moebius transformation that meets your requirements. That is, for any $a>0$, $$M_2(z) = a(M(z)-\sqrt{2}(1+i)) + \sqrt{2}(1+i)$$ will be another Moebius transformation that meets your requirements.

Answer 2

$p = \sqrt{2} + i \sqrt{2}$ is a rather inconvenient point, so start by mapping it to $\infty$. The transformation $h(z) = 1/(z-p)$ takes $p$ to $\infty$, takes the circle $|z|=2$ (which passes through $p$, $-p$ and $2$) to a straight line through $h(-p)$ and $h(2)$, and takes the line $\text{Re}(z) = \text{Im}(z)$ (which passes through $p$, $-p$ and $\infty$) to a line that passes through $h(-p)$ and $0$. $g = h \circ M \circ h^{-1}$ should be a Möbius transformation that fixes $\infty$ (thus is of the form $g(w) = a w + b$) and maps a certain half-plane to another half-plane...