A question on Conway's Complex Functions of One Variable asks: Find necessary and sufficient conditions for a Mobius transformation $T(z)=\frac{az+b}{cz+d}$ to map the unit circle to itself. So if $\gamma$ is a circle, $T(\gamma)=\gamma$.

I've worked out the necessary conditions. Namely, if $T(\gamma)=\gamma$, then

1) $|a|^2+|b|^2=|c|^2+|d|^2$

2) $a\bar{b}=\bar{d}c$

3) $\bar{a}b=d\bar{c}$

How does one go about showing sufficiency? Should I simply assume conditions 1),2) and 3) and try to prove that $T(\gamma)=\gamma$? If so I can simply claim that all the implications I used to get these conditions also work backwards. Or just show that $|\frac{az+b}{cz+d}|=1$ by these conditions, which is rather simple. Is that all there is to this? I just wish the whole "necessary and sufficient" language was scrapped for some direct notation.

As an aside, I'm wondering if I'm using the words "necessary" and "sufficient" correctly in this context. Is what I showed in the first part the necessary conditions (that's what makes sense to me semantically, because they are necessary once I've assumed the map), or are they the sufficient conditions?

Any help is much appreciated.

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    $\begingroup$ Conditions 2 and 3 are the same essentially. $\endgroup$ – Matt Samuel Mar 20 '16 at 2:19
  • $\begingroup$ @MattSamuel Right, they are one conjugation apart, thanks for pointing that out. $\endgroup$ – Mike Mar 20 '16 at 2:20
  • $\begingroup$ Related: math.stackexchange.com/questions/209308/… $\endgroup$ – Alan Muniz Mar 20 '16 at 16:56
  • $\begingroup$ Thx for ur link, that thread is botched because the op asked about the unit disk initially, so people gave him answers to that question, and then he changed it to a circle, some people even answer with the typical mobius transformation that maps the unit disk to itself in an attempt to answer his circle question... that hardly gives necessary and sufficient conditions to my problem. $\endgroup$ – Mike Mar 20 '16 at 17:01

The map $T(z) = \frac{az+b}{cz+d}$ sends the unit circle to itself if and only if for any $\zeta$ in the circle, $|T(\zeta)|=1$. Now you just have to translate this into conditions on the coefficients.

$|T(\zeta)|=1$ is equivalent to $$ |a|^2 + |b|^2 + 2Re(a\bar{b} \zeta) = |c|^2 +|d|^2+ 2Re(c\bar{d}\zeta) $$

Since $T$ is determined by the image of three points, just evaluate the equation above for three distinct values of $\zeta$ and you'll get the necessary and sufficient conditions. For $\zeta = 1, i, -1$, you'll get the conditions 1) and 2).

  • $\begingroup$ Thank you for your answer. As my question suggested, I've followed this process precisely. My real question is, why does this process give me the necessary AND sufficient conditions? Isn't this an if and only if statement? We started off supposing $|T(z)|=1$, and we found the restraints on $a,b,c$ and $d$ to get the necessary conditions. Now, shouldn't we work backwards to show that these conditions are also sufficient? Or perhaps they are not and we require a few more conditions for sufficiency. And of course I might be using the words necessary and sufficient backwards, but that's fixable. $\endgroup$ – Mike Mar 20 '16 at 18:11
  • $\begingroup$ Untill we achieve the equation, every thing is "if and only if". And the equation is true for every $\zeta$ if and only if it is valid for three distinct values, thats the point. $\endgroup$ – Alan Muniz Mar 20 '16 at 18:14
  • $\begingroup$ If we change the points we'll get equivalent conditions $\endgroup$ – Alan Muniz Mar 20 '16 at 18:15
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    $\begingroup$ Ah I see what you mean, mobius transformations are uniquely determined by their actions on three points. $\endgroup$ – Mike Mar 20 '16 at 18:15
  • $\begingroup$ Correct me if I'm wrong, but I don't believe that it is that fact alone which is used for the converse. We also have to recall that the image of a circle under a Mobius transformation is a circle. So the image of the unit circle is going to be a circle. The condition being satisfied for $1,i,-1$ means that $|T(1)|=|T(i)|=|T(-1)|=1$ i.e., the images of $1,i,-1$ lie in the unit circle. Hence the image must be the unit circle. $\endgroup$ – Hrit Roy Feb 26 '19 at 15:48

here we have a unit circle $\gamma$. we want to find the sufficient and necessary condition such that $T(\gamma)=\gamma$.

we know that for every point $z$ in the unit circle, $\bar{z}$ is also in it. so $|\bar{z}|=1$ so we have $|z^2|=1$ and $z\bar{z}=1$.

$|T(\gamma)|$ must be also equal to $1$. so for every $z$, $T(z)\bar{T(z)}=1$. after solving and simplify, we have:


it must be just like $z\bar{z}-1=0$.

so :





this yields the sufficient condition $|a|^2+|b|^2=|c|^2+|d|^2 $ $(*)$

to finding the necessary condition we assume that $c=\lambda\bar{b}$ and this yields $d=\frac{\bar{a}}{\lambda}$. from inserting $d$ and $c$ in $(*)$, $|\lambda|$ must be equal to $1$. and if we take $\lambda=e^{i\alpha}$, $T(z)=e^{i\alpha}\frac{az+b}{\bar{b}z+\bar{c}}$ is the necessary condition.


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