Image of a Domain under Möbius transformation

Question

I want to determine the image of the open Triangle with cornerpoints 0,1,i under the map f(z)=$\frac { 1 }{ z } $. We have that the points 0,1,i are mapping to $0\rightarrow \infty \quad 1\rightarrow 1\quad i\rightarrow -i$. So since 0 maps to infinity we are mapping the given Domain to line where all z values are mapped to -Im(z). Can u help me to continue, how to i find the domain im looking for.

Answer 1

We msut look at the lines.

• The line through $0$ and $1$ (and $\infty$) turns into the line through ($\infty$ and) $1$ and $0$ - so the same line.
• The line through $0$ and $i$ (and $\infty$) turns into the line through ($\infty$ and) $-i$ and $0$ - so the same line.
• The line through $1$ and $i$ (and $\infty$) turns into the circle through $1$ and $-i$ and $0$. This is the circle with centre $\frac{1-i}{2}$ and radius $\frac{\sqrt 2}2$.

Remains to see, which of the regions of the plane these lines and circle determine is the correct image. It must of course be that region tha tcontains $f(10^{-1000}+10^{-1000}i)$ ...