How to show that the map if $ad-bc\ne0$ then the map given by $z\mapsto\dfrac{az+b}{cz+d}$ is non-constant where $a,b,c,d\in\mathbb C$.

I know that the mapping together with the condition is a Möbius Transformation. But I can't show the condition ensures that the map is non-constant.

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    $\begingroup$ … map is non-constant? $\endgroup$
    – k.stm
    May 5, 2013 at 10:20

5 Answers 5


If the map is constant then for all $z$ and for some constant $k$$$\frac{az+b}{cz+d}=k\Rightarrow (a-kc)z+(b-kd)=0$$ and this implies $a=kc$ and $b=kd$. Therefore $a/c=b/d$, which is a contradiction.


1) This function is holomorphic on $\mathbb{C}\setminus\{-\frac dc\}$.
2) If a holomorphic function $f$ is constant then $f'=$...
3) If $f(z)=\frac{az+b}{cz+d}$ then $f'=$..., so...


The map $z ↦ \dfrac{dz - b}{-cz + a}$ gives an actually an inverse on $a ≠ cz$.


If $f$ is your rational map, $f(0)=\frac{b}{d}$ and $f(1)=\frac{a+b}{c+d}$.
Your bright little sister who has just learned about fractions will tell you that $\frac{b}{d}=\frac{a+b}{c+d}$ is exactly equivalent to $ad-bc=0$ .
Since your hypothesis excludes this equality, you must have $f(0)\neq f(1)$, so that $f$ is not constant.

[In case $d=0$ or $c+d=0$, the argument must be slightly modified or, better, slightly reinterpreted by considering the value $\infty$ ]


If you let $A$ be the 2-by-2 matrix with entries $[a,b;c,d]$, then one nice visual way to think of Möbius transformations

$f(z) = \frac{az+b}{cz+d}$

is as $A$ applied to the affine subspace $\{ (z,1)| z\in\mathbb{C}\}\subset\mathbb{C}^2$, and then projected along the line going through the origin back to that subspace (and finally picking out the first coordinate, the second being just 1):

$f(z) = g(A\cdot (z,1)^T))$


$g((z,w)^T) = z/w$.

So when $ad-bc\neq 0$, then $A$ is by definition nonsingular, and there's no way for $A$ to move the subspace $\{(z,1)\}$ on a line through the origin, which would be required to make the projection back onto that subspace a constant.

This is certainly a bit more work than direct ways, but realizing that Möbius transformations can be thought of in this way makes several questions regarding them intuitively clear if you like to think geometrically (also it makes the $ad-bc\neq0$ condition not out-of-the-blue).


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