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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!

MJD
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user137684
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2 Answers2

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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.

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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.

Mark McClure
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  • Why did you select this form to start with? Can your final $M(z)$ be of the form $\frac{az+b}{cz+d}$? It appears to, but I don't understand where you got your initial $f(z)$ from. Can you explain? thx – user137684 Jan 08 '15 at 17:43
  • @user137684 From experience, I knew that I could map the circle $|z|=2$ to a line and that the interior of the circle would map to a half plane bounded by that line; that is the role of $T(z)$. Once we have that, it is a simple matter of rotating the line onto the line $Re(z)=Im(z)$ (via multiplication by $\rho=e^{3\pi i/4}$ and taking care to make sure that zero maps to the correct side) and then shifting along that line to get the right fixed point. The result is Moebius transformation since composition of Moebius transformations yields another. – Mark McClure Jan 08 '15 at 17:49
  • Ok great thank you. I am now having trouble finding where the line ends up after $T(z)$. I understand that we take $\frac{1}{z-2}$ as the circle inversion, but then have to take the complex conjugate to get zero on the correct side of the line, but what is the circle we are doing circle inversion in? That is the only way I know to find the mapping of the curve. – user137684 Jan 08 '15 at 21:17
  • @user137684 I expanded and illustrated my response a bit. I used $T(z)=(z+2)/(z-2)$ in my explanation. You can certainly use $1/(z-2)$, though your rotation will be different. A key fact is that all these transformations are themselves Mobius transformations, so their compositions are as well. You cannot use conjugation, as it is not a Mobius transformation - it's not even analytic! But you don't need to. – Mark McClure Jan 09 '15 at 03:26
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$p = \sqrt{2} + i \sqrt{2}$ 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...

Robert Israel
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