# Is the complex conjugation map a Mobius transformation?

I have been asked to prove whether the complex conjugation map $$z\mapsto \bar{z}$$ is a Mobius transformation.

My solution was: Suppose $$\bar{z}\in \mathcal{M}$$, so that we can write $$\bar{z} = \frac{az+b}{cz+d} = f(z)$$ for some $$f\in\mathcal{M}$$.

As the equality must hold for all $$z\in\mathbb{C}\cup\{\infty\}$$, we can consider particular values of $$z$$ to determine the coefficients.

$$z=0 \implies b=0$$

$$z=\infty \implies c=0$$

$$z=1 \implies a=d$$

So this means our map must be $$f(z)=z = \bar{z}$$. But $$z=1+i \implies \bar{z} = 1-i$$ and so the two maps do not agree on $$z=1+i$$. We conclude that $$\bar{z}$$ is not a Mobius transformation.

Is this okay?

• That's good, but you are making an unecessary assumption about the meaning of $\overline{\infty}$: you must have $c = 0$ (and $d\neq 0$), because otherwise you could find a finite $z$ such that $\overline{z}$ is "undefined" or "not in $\Bbb{C}$". Jun 3 at 19:46

Another way is to consider points $$z$$ on the circle $$|z| = r \gt 0$$. Then $$\bar z = \dfrac{r^2}{z} = f(z)$$, and comparing coefficients gives $$a = d = 0, \dfrac{b}{c} = r^2$$. But this cannot hold true for all $$r \gt 0$$.
Sure, that’s fine (as long as you know why the arithmetic with $$\infty$$ works out). I would have simply said the Mobius transformations are holomorphic (in the usual sense, away from the zero of the denominator, i.e at all but one point of the complex plane), whereas the function $$f(z)=\overline{z}$$ is nowhere differentiable. So, complex conjugation cannot possibly be a Mobius transformation.