A Möbius transformation is given by


with parameters $a$, $b$, $c$, and $d$. The Wikipedia article provides rules for finding these parameters based on three points $z_1$, $z_2$, and $z_3$ and their images $w_1$, $w_2$, and $w_3$. It is my goal to understand how we can derive the equations which yield the parameters.

Möbius transformations preserve the cross-ratio, so I assume we start with the cross-ratios of the original points and their images: $$(z,z_1;z_2,z_3)=(f(z),w_1;w_2,w_3)$$ which can be reformulated as


I imagine the solution is obtained by reformulating this equation above somehow to solve for $f(z)$. But how is this done? I could not find a proper tutorial for this online - most tutorials I find plug in specific points at this stage, but I would like to learn how the general approach is derived.

  • $\begingroup$ Did you try to solve the last equation for $f(z)$? It is not that difficult. Start by multiplying with $(w_1-w_2)/(w_1-w_3)$. $\endgroup$ – Martin R Aug 29 '20 at 20:36
  • $\begingroup$ Yes, I did. In that case, I end up with $\frac{f(z)-w_2}{f(z)-w_3}$ on the RHS, and I don't know how to solve this for $f(z)$. $\endgroup$ – J.Galt Aug 29 '20 at 20:42
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    $\begingroup$ Then multiply with $f(z)-w_3$ and collect the terms containing $f(z)$ on one side ... $\endgroup$ – Martin R Aug 29 '20 at 20:44
  • $\begingroup$ Hm, thank you for the tip! I also tried this before but couldn't cleanly separate the $f(z)$ from the $z$. I must have made an error - I'll revisit my calculations and see where the error is. $\endgroup$ – J.Galt Aug 29 '20 at 20:46
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    $\begingroup$ You can also write the right-hand side as a composition of simple transformations and reverse these in opposite order. $\endgroup$ – Martin R Aug 29 '20 at 20:48

multiply up



  • $A = (w_1-w_2)(z-z_2)(z_1-z_3)$
  • $B = (w_1-w_3)(z_1-z_2)(z-z_3)$

$$(f(z)-w_3)\cdot A=(f(z)-w_2)\cdot B$$

multiply out $$f(z) \cdot (A-B)=w_3\cdot A-w_2\cdot B$$

multiply out

$$f(z) =\frac{w_3\cdot A-w_2\cdot B}{A-B}$$

should work as long as $A \not = B$, infinities may need to be handled separately.


Unless you have an unusual memory, and/or are somehow required to do this, I think there's little reason to have in mind "the formula" or any "clever changes of variables"... to determine the coefficients of the linear fractional (Mobius) transformation you want.

Rather, it is possible to derive the expression in a natural fashion (thus dodging the brittleness of memory-of-formulas...). Namely, first move $z_1$ to $w_1$ by your choice of simple linear fractional transformation, e.g., either a complex multiplication or a complex addition. Keep track of where the other points go. Then choose a l.f.t. fixing the first desired image $w_1$, and moving the new $z_2$ to $w_2$. And similarly for (the new-new) $z_3$ to $w_3$.

A minor variant of this, which is probably what I'd use, since I'd have to try to figure out those point-stabilizing groups, is to map the $z_i$'s to some canonical special points, such as $0$, $1$, $\infty$, whose isotropy groups are easy to understand... If $g$ sends $z_1,z_2,z_3$ to $0,1,\infty$, and $h$ sends $w_1,w_2,w_3$ to the same, then $h^{-1}g$ sends $z_i$ to $w_i$.

(It is cool to know that there is an explicit invariant, the cross-ratio, but I'd not use it as a computational device... especially since one needs to be able to accurately recall it. One thing I've learned, even though my memory seems to be ok, is that raw memory without the robustness of an easily memorable explanation, is dangerously fragile... Like remembering phone numbers...)


Let's take as an example probably the most famous Mobius transformation, the Cayley transformation. It is determined by the fact that it takes $i\to0, -i\to\infty$ and $0\to-1$.

Let's find the formula for the transformation, given this information. So $f(z)=\dfrac{az+b}{cz+d}$ and we get the three equations: $\dfrac{ai+b}{ci+d}=0, \dfrac{-ai+b}{-ci+d}=\infty$ and $\dfrac bd=-1$.

From this, a little algebra leads to $f(z)=\dfrac{z-i}{z+i}$.

Since the algebra seems to be the part you're asking about, let's do it. We get $\dfrac{ai+b}{ci-b}=0, \dfrac{-ai+b}{-ci-b}=\infty \implies f(z)=\dfrac{biz+b}{biz-b}=\dfrac{z-i}{z+i}$.


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