Conformal map of unit disk to itself

[Dromepalin]

Homework Statement

This problem is an already solved one in Marsden and Hoffman's Basic Complex Analysis, but I can't seem to work out the last step.
Here's the problem:

Suppose a,b,c,d are real and ad-bc>0. Then show that $$T(z) = \frac{az+b}{cz+d}$$ leaves the upper half plane invariant. Show any conformal map of the upper half plane is of this form.

Homework Equations

The Solution to the second half:
Suppose T maps the upper half plane conformally to itself. Then
$$W(z) = \frac{T(\frac{1}{i}*\frac{z+1}{z-1})-i}{T(\frac{1}{i}*\frac{z+1}{z-1})+i}$$
is a conformal map of the unit disk to itself. Now W has the form:
$$W(z) = e^{i\theta}\frac{z-z_{0}}{1-\overline{z_{0}}z}$$

and so $$e^{i\theta}\frac{\frac{w-i}{w+i}-z_{0}}{1-\overline{z_{0}}\frac{w-i}{w+i}} =\frac{T(w)-i}{T(w)+i}$$

Solving for T(w): <=== This is the part I don't understand!
$$T(w) =\frac{w*Re[(1-z_{0})e^{i\theta/2}] - Im[(1+z_{0})e^{i\theta/2}]}{w*Im[(1-z_{0})e^{i\theta/2}] + Re[(1+z_{0})e^{i\theta/2}]}$$

The Attempt at a Solution

How on Earth did the author get that value for T(w). I have tried pages of mindless algebraic manipulation to no avail.

 

[lippyka]

I think it has something to do with the fact that T(w) is a fraction, and so the numerator and denominator can't be combined. I am also aware of the fact that Re[z] = \frac{z+\overline{z}}{2} and Im[z] = \frac{z-\overline{z}}{2i}. Any help would be much appreciated!