Map $w(z)=\frac{1}{1+z^2}$ for $0<|z|<1$ is Conformal

Question

Question Map from punctured disc $$D^*:=\lbrace z\in\mathbb{C}:0<|z|<1\rbrace$$ to a domain $P$ $$P:=\lbrace w\in\mathbb{C}:1/2<\Re(w), w\neq 1\rbrace$$ defined as $$w:D^* \to P $$ $$w(z)=\frac{1}{1+z^2} $$ is Conformal.

Attempt $$w:D^* \to P $$ $$w(z)=\frac{1}{1+z^2} $$ $w(z)$ is analytic for $z\in D^*$ $$w'(z)= \frac{-2z}{(1+z^2)^2}$$ $$w'(z)\neq 0 \ \forall \ z\in D^*$$ Also $D^*$ is connected. Hence the map is conformal. Any help is appreciated!

Answer 1

The best way to look at this, in my opinion, is to look at your map as the composite of 3 functions:

• First, the square map $z \mapsto z^2$,
• Next, the translation $z \mapsto z + 1$,
• Then, the inverse map $z \mapsto 1/z$.

The derivative of the first one is multiplication by $2$, which is a homothety, and it is conformal. The derivative of the second one is multiplication by $1$, even better. Since a composite of conformal maps is conformal, you only need to know that the inverse map is. And it is, since it is $\mathbb C$-derivable: its differential at a point $z$ is multiplication by the non-trivial complex number $-1/z^2$, and multiplication by a non-trivial complex number is a homothety (which is the reason why holomorphic maps with non-vanishing derivative are conformal).