Activity 1 (Commented)

Using Maple, verify that  the bilinear transformation   is a one-to-one transformation from the unit disk    onto the upper half plane   .

> with(plots):

Warning, the name changecoords has been redefined

> f:='f': s:='s': w:='w': z:='z':

f := z -> I*(1-z)/(1+z):

`w = s(z) ` = f(z);

To show   is  one-to-one,  find the inverse function:

> `z ` = solve(w=f(z),z);

To verify that    is onto we use a graph which is not conclusive:

> f:='f': z:='z':

f := z -> I*(1-z)/(1+z):

`s(z) ` = f(z);

conformal(f(Re(z)*exp(I*Im(z))), z=0.01..1+I*2*Pi,

  title=`w = i(1-z)/(1+z)`,

  labels=[`u`,`v`], tickmarks=[7,4],

  grid=[25,25], numxy=[25,25],

  scaling=constrained,

  view=[-3..3,0..3]);

Activity 2

Now prove mathematically that  the bilinear transformation   is a one-to-one transformation from the unit disk    onto the upper half plane   .