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From my research, I have figured out that this is a Möbius transformation. The respective wiki page helped me understand a bit more, however I can't figure out how to obtain the image.

So lets talk about what I do know. Well we are describing the set of all values $f(z)$ where $|z| < 1$. I also know that any three points.

I also know that

Given a set of three distinct points z1, z2, z3 on the Riemann sphere and a second set of distinct points w1, w2, w3, there exists precisely one Möbius transformation f(z) which maps the zs to the ws

and from help from another forum I got the following:

To decide which part is the image of the interior |z| < 1 of the disc, figure out which point f sends to infinity.

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  • $\begingroup$ you accidentally removed the expression for your function $\endgroup$
    – Zarrax
    Sep 29, 2011 at 18:19

3 Answers 3

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Hint. The inverse of $w=\dfrac{3z+i}{-iz+3}$ is $z=\dfrac{3w-i}{iw+3}$. So $|z|<1$ is equivalent to $|3w-i|<|iw+3|$, or $3|w-\dfrac{i}{3}|<|i||w-3i|=|w-3i|$. One can find the desired image by considering the Apollonius' circle in the complex plane ($w$-plane) or by squaring both sides and using $|z|^2=z\overline{z}$.

Edit: To elaborate on the 'square both sides and use $|z|^2=z\overline{z}$' part: $$\begin{align*} |3w-i|^2&<|iw+3|^2\\ (3w-i)(3\overline{w}+i)&<(iw+3)(-i\overline{w}+3)\\ 9w\overline{w}+3iw-3i\overline{w}+1&<w\overline{w}+3iw-3i\overline{w}+9\\ 8w\overline{w}&<8\\ |w|^2&<1\\ |w|&<1 \end{align*}$$

Hence the image is again the open unit disk.

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  • $\begingroup$ +1 for the hint containing every indication needed to write a full solution. $\endgroup$
    – Did
    Sep 29, 2011 at 7:26
  • $\begingroup$ I get this but I have a problem. $w$ is exactly equal to $z$ so obviosuly it will end up with $|z| < 1$. We need to show what the image of f is.. not the image of the inverse function. Am I making sense? $\endgroup$
    – Tyler Hilton
    Sep 29, 2011 at 20:47
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    $\begingroup$ $w=f(z)=(3z+i)/(-iz+3)$ is obviously not equal to $z$. The above is a common procedure for finding the image of a function. Do you understand that $|z|<1$ is equivalent to $|3w-i|<|iw+3|$? Which means, if $|z|<1$, its image $w=f(z)$ must satisfy $|3w-i|<|iw+3|$; and if $w$ satisfies $|3w-i|<|iw+3|$, it must be the image of some $z$ satisfying $|z|<1$. This shows that $\{w\in \mathbb{C}\mid|3w-i|<|iw+3|\}$ is the desired image. $\endgroup$
    – pharmine
    Sep 30, 2011 at 3:13
  • $\begingroup$ thankyou, however i am still a littttle bit confused. Okay, I understand that z is equivalent to the inverse function. So basically instead of saying $|z| < 1$ I can say inverse < 1. Now lets pick an element from the range (ie, $w = f(z)$). We can say that since $|z| < 1$ then the image $w$ and its magnitude $|w| < 1$. but z is equivalent to the inverse function and so we plug w in there. What i am confused about is how are we certain that the image has magnitude less then 1. Maybe the function takes it outside the unit disk, or takes it to infinity. $\endgroup$
    – Tyler Hilton
    Sep 30, 2011 at 16:05
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There are several ways to do these, but most of them are facilitated by sort of knowing the answer. Recall that fractilinear transformations preserve circilinearity, i.e. circles and lines get mapped to circles and lines (NOT respectively - perhaps a circle to a line or vice versa). And a line is a circle that goes through infinity (really - it's a good way to think of these things).

So if I were you, I would get an idea for where the circle goes by looking at the boundaries of the complex unit cube, or perhaps the four easy coordinates (the purely real and the purely imaginary) of the unit circle (or both), and see where they go. This gives a good idea.

Also, I would note that the new 'infinity point' is when the denominator is zero.

Does that give you a good start?

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  • $\begingroup$ I had to google "fractilinear" and "circilinearity" to see if I was missing those awesome terms all these months or if you just made them up. $\endgroup$
    – anon
    Sep 29, 2011 at 4:02
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    $\begingroup$ @anon: You know, reading this made me google them. In my complex class, we used 'circilinear' all the time, but Google thinks I made them up. It's okay - now everyone will think I'm witty. $\endgroup$
    – davidlowryduda ♦
    Sep 29, 2011 at 4:05
  • $\begingroup$ It does, but i lost you at the second paragraph. My boundaries on the unit circle is of radius 1 at the x (real) and y(im) axis right? But what do you mean by seeing where they go? should I plug 1 into my formula, then 1 + a small amount and 1 - a small amount? $\endgroup$
    – Tyler Hilton
    Sep 29, 2011 at 4:06
  • $\begingroup$ @Tyler: I'm saying that 1,0; 0,1; -1,0; 0,-1 are all on the complex circle (in real, complex form). So their images will all be on the same line/circle too. $\endgroup$
    – davidlowryduda ♦
    Sep 29, 2011 at 4:09
  • $\begingroup$ Ok, thanks. One thing I need to clear: What do you mean by an image? Also I picked point 0.5, plugged into $f(0.5)$ = $\frac{35}{4}i$. If i pick $z=1+0i$ I get $\frac{3+i}{3-i}$. I dont seem to know how to interpret this. Maybe an image would help? $\endgroup$
    – Tyler Hilton
    Sep 29, 2011 at 4:15
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If you plug in $z = e^{i\theta}$ into your linear fractional transformation you get $${3e^{i\theta} + i \over -ie^{i\theta} + 3}$$ This is the same as $$e^{i\theta} {3 + ie^{-i\theta} \over 3 - ie^{i\theta}}$$ The numerator and denominators of the fraction are complex conjugates of each other, so it has magnitude $1$, as does $e^{i\theta}$. So their product also has norm $1$ for all $\theta$ and therefore the linear fractional transformation takes the unit circle to itself.

In general when it takes the unit circle to itself you can use factorizations this way to show it.

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  • $\begingroup$ why plug that in, you can acheive the same thing with z, no? $\endgroup$
    – Tyler Hilton
    Sep 30, 2011 at 0:16
  • $\begingroup$ Nice argument, but I think it is not immediately clear whether the transformation takes the unit circle to "the whole of" the unit circle. $\endgroup$
    – pharmine
    Sep 30, 2011 at 7:45
  • $\begingroup$ Whats the difference between this and the one you have? $\endgroup$
    – Tyler Hilton
    Sep 30, 2011 at 16:20
  • $\begingroup$ @pharmine the inverse function can be directly computed (as you did) and the same argument works for the inverse function: ${\displaystyle {3e^{i\theta} - i \over ie^{i\theta} + 3} = e^{i\theta}{3 - ie^{-i\theta} \over 3 + ie^{i\theta}}}$. Once again the numerator and denominator are complex conjugates of each other. $\endgroup$
    – Zarrax
    Oct 1, 2011 at 14:57
  • $\begingroup$ @TylerHilton the idea is, you want to show the unit circle goes to itself.. anything on the circle is of the form $e^{i\theta}$. So plug that in and try to show the magnitude of the result is $1$. $\endgroup$
    – Zarrax
    Oct 1, 2011 at 15:00

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