I need to find a Möbius transform $\phi$ that satisfies :
$(1) \phi(i)=-i$
$(2) arg(\phi'(i))=-\pi/2$
$(3) \phi(\{Imz>0\})=\{Imz<0\}$
I'd be thankful for some hints.
Thanks.
I need to find a Möbius transform $\phi$ that satisfies :
$(1) \phi(i)=-i$
$(2) arg(\phi'(i))=-\pi/2$
$(3) \phi(\{Imz>0\})=\{Imz<0\}$
I'd be thankful for some hints.
Thanks.
Since $\varphi(\{Imz>0\})=\{Imz<0\}$, you know that $\varphi(\mathbb{R})=\mathbb{R}$. which happens if and only if
$\varphi(z)=\frac{az+b}{cz+d},\hspace{10pt}a,b,c,d\in\mathbb{R}$ (Can you prove this?)
Now, observe that if $\varphi(\mathbb{R})=\mathbb{R}$ and $\varphi(i)=-i$ then $\varphi(\{Imz>0\})=\{Imz<0\}$, so it is enough to find $\varphi$ as I mentioned above with $\varphi(i)=-i$ and $\arg(\varphi'(i))=-\pi/2$.
Remember that $\arg(\varphi'(i))=-\pi/2$ if and only if $\frac{ad-bc}{(ci+d)^2}=\varphi'(i)=\alpha i$, $\alpha<0$.
This gives you some equations on $a,b,c,d$ and you can find at least one sample.