Area of a holomorphic function's range on the unit disk
Suppose , where D(0,1) is the open unit disc, f is one-to-one in D(0,1), , and . Prove that the area of is .
Here are my partial solutions:
The Jacobian of f is
(I worked this out, but I'm not showing it here because I'm fairly confident it is correct)
The area of is
Using our change of variables
We use the uniform convergence of the power series to swap integration and summation...
Here I made a mistake (we can't just take the square in term by term), but it seems to have led to the right answer.
Now we switch to polar coordinates via .
This gives .
I don't understand how my incorrect simplification of the square of the sum led to this correct result.
When I have tried to proceed through properly applying the square, I don't know how to simplify to a result I can evaluate.
I'm not sure where I am/should be using the one-to-oneness of .
Any help/hint is greatly appreciated.