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Question: Find the bilinear transformation which carries the circle $|z|=3$ into $|z-1|=1$, the point $3+3i$ into $1$ and the point $3$ into $0$.

My Attempt: First, I've done problems like this before, but I ran into something at the beginning, so I want to first write my proof, then mention what I ran into and hope to have my solution verified and to see if I can get an explanation on why I had that problem and what it means. So here goes:

Let $f(z)=w$ we such a transformation. We have that $f(3+3i)=1$ and $f(3)=0$. Now, we see that, using $z^*=\frac{R^2}{\bar z-\bar a}+a$, that $3+3i$ is symmetric to $3-3i$ with respect to $|z|=1$ and that $1$ is symmetric to $\infty$ with respect to $|z-1|=1$. Thus, we have $f(3-3i)=\infty$. So, using the cross ratio, we have $(w,1,0,\infty)=(z,3+3i,3,3-3i)$, and so we have $\frac{w-0}{w-\infty}\frac{1-\infty}{1-0}=\frac{z-3}{z-(3-3i)}\frac{3+3i-(3-3i)}{3+3i-3}$. Going through the calculation, we get that $w=\frac{(z-3)2i}{z-3+3i}$.

So, first, does this solution look correct? Next, when I was trying to find that "last point", I was running into an issue using $f(3)=0$. I kept getting $3$ is symmetric to $3$ with respect to $|z|=3$ and $0$ is symmetric to $0$ with respect to $|z-1|=1$, so I was just getting a transformation value that I already knew. Why? Or, should I, of course, have used the point with the nonzero imaginary part? Any help is greatly apprecaited! Thank you.

User7238
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1 Answers1

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The transformation required is $f(z)=\frac{(z-3)(1-i)}{2z-3-3i}$

One can find it by noting that $g(w)=f(3/w)-1, |w| \le 1$ is a unit disc automorphism that carries $\frac{1-i}{2} \to 0$ and $1 \to -1$ hence it is $g(w)=\alpha \frac{w-\frac{1-i}{2}}{1-\frac{1+i}{2}w}, |\alpha|=1$ and putting $w=1$ gives $\alpha=i$, so giving $g(w)= i \frac{w-\frac{1-i}{2}}{1-\frac{1+i}{2}w}$ and $f(z)=g(3/z)+1$ which when explicited leads to the answer above.

(an easy check shows that the required values are attained and then $f(3i)=1-i$ and $f(-3)=\frac{2-4i}{5}$ which are clearly on $|z-1|=1$ together with $f(3)=0$ and since a circle is uniquely determined by three points and $f$ sends circles to circles, shows that $f(|z|=3)$ is indeed $|z-1|=1$)

Conrad
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  • Thanks, Conrad! I am just really wondering WHY I was running into that issue when finding symmetry points. Also, since my transformation doesn't seem to match yours, I am wondering if I have a silly algebra mistake somewhere, or what it could be. – User7238 Aug 11 '21 at 14:10
  • Should be $(1-i)/6$ symmetric with $3+3i$ not $3-3i$ as the product must be $1$ – Conrad Aug 11 '21 at 14:38
  • maybe I am not seeing something, but I see I made a mistake in my original calculation, but I am still not getting $(1-i)/6$... Using $z^{\star}=R^2/(\bar{z}-\bar{a})+a$, I get $z^{\star}=\frac{9}{3-3i}=\frac{3+3i}{2}$. So, So $3+3i$ is symmetric to $\frac{3+3i}{2}$ with respect to $C_1:=|z|=3$. Similarly, we get that $1$ is symmetric to $\infty$ with respect to $|z-1|=1$, so we should have $f(\frac{3+3i}{2})=\infty$, right? – User7238 Aug 15 '21 at 01:49
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    I guess you are right and the denominator has indeed $\frac{3+3i}{2}$ as a root as it is $2z-3-3i$ (this shows how easy is to make mistakes if you stray from the path that is familiar - for me that is using the unit circle automorphisms as noted in the answer) – Conrad Aug 15 '21 at 01:59
  • Thank you, @Conrad! I do appreciate you showing what you did, though. I always like having another "tool" in case I encounter such problems :) – User7238 Aug 15 '21 at 11:55
  • Hi Conrad! I was revisiting this problem, and I just can't get it to come out right. After I found my points of symmetry, I am using the cross ration $(w,\infty,1,0)=(z,\frac{3+3i}{2},3+3i,3)$, and I keep getting $\frac{zi-z+6}{-z-zi+3+3i}$. I figured it was poor algebra on my behalf, but after doing it 4 or 5 times, unless its REALLY poor algebra on my behalf, I keep getting the same answer. I am not saying that I am disagreeing with your original answer, but I am not sure why this answer isn't matching... – User7238 Dec 03 '21 at 02:18
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    Using your numbers one has $\frac{w-0}{w-\infty}\frac{1-\infty}{1-0}=\frac{z-3}{z-(3+3i)/2}\frac{3+3i-(3+3i)/2}{3+3i-3}$ and that should simplify to my original answer when you move the $2$ in the $z$ denominator up as well as divide by $3i$ in the free denominator – Conrad Dec 05 '21 at 20:39
  • Ah, okay, so, just to be clear, when using the cross ratio, if we have $(z_1,z_2,z_3,z_4)=(w_1,w_2,w_3,w_4)$, you are using $\frac{w_1-w_4}{w_1-w_2}\frac{w_3-w_2}{w_3-w_4}=$(same corresponding stuff on $z$ side) because it is convenient because your left hand side is just $w$. In other words, we can write the cross ratio in a bunch of different ways, as long as it is consistent on both sides... for instance, it doesn't matter if we write it the way you did, or like $\frac{w_1-w_2}{w_1-w_4}{w_3-w_4}{w_3-w_2}$.... in any case, it looks like I am just doing bad algebra – User7238 Dec 05 '21 at 20:48
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    I followed the formula from your post and it worked well - it is simplest cross ratio and as a general principle, using simplest formula allows for least mistakes – Conrad Dec 05 '21 at 23:48
  • Thank you, @Conrad! As always, I really appreciate your insight and help!! – User7238 Dec 05 '21 at 23:49