Given $w(z)=\frac{i-z}{i+z}$.
Find the map w=f(z) of the part of the plane defined by inequalities:
$|z|>1$ and $Im(z)>Re(z)$
so far:
$|z|>1$ is this area
from $Im(z)>Re(z)$ => $y>x$
and together is
first thing to do is to express z as a function of w
$w=\frac{i-z}{i+z} <=> wi +wz=i-z <=> ... z=\frac{i(1-w)}{1+w}$
$w=u+iv$
$z=\frac{i(1-u-vi)}{1+u + vi} ..... <=> z=\frac{-2 v}{(1+u)^2+v^2}+i (\frac{-1}{(1+u)^2+v^2}+\frac{u^2}{(1+u)^2+v^2} +\frac{v^2}{(1+u)^2+v^2})$
now my question is what to do next?
here some similar example
given is x=1 and x=2 (area between). Mapping is defined with $w=\frac{z+2}{z}$.
Answer:
x=1 is mapped in $(u-2)^2+v^2=1$
x=2 is mapped in $(u-\frac{3}{2})^2+v^2=\frac{1}{4}$
now area between x=1 and x=2 is mapped into plane between this two circles
First note that $w(z) =\frac{i-z}{i+z}$ is a Mobius transformation (linear fractional transformation). Such functions take circles or lines to circles or.lines.so,this saves a lot of work since you look for the image of 3 points and determine theimage of line or circle.
Now ,the image of unit circle is imaginary line,interior of it goes.to right half-plane and exterior goes to left halfplane.Theline in question is mapped to a circle and upper part goes to the inside of the disk.Finally,take the intersection to get the image of the shaded region.