# Proof regarding Stein's and Sharachi's proof of the unit circle mapping onto the upper half plane - complex analysis

I am currently working with Stein's and Shakarchi's Complex Analysis and trying to comprehend the proofs of certain theorems. But unfortunately i can't get behind the equation used in the proof of Theorem 1.2:

I understand why we are computing $$\Im(G(w))$$ but i simply do not understand why it holds that $$\Im(G(w)) = \Re(G(u+iv))$$ even though i fully understand using $$w = u+iv$$ in order to compute both real part and imaginary part.

But the author simply computes the real part and claims that since $$\Re(G(u+iv)) > 0$$ it holds that $$\Im(G(w)) > 0$$ ?

I tried to solve it myself but i always end up having

$$\Re(G(u+iv)) = \frac{1-u^2-v^2}{(1-u)^2+v^2}$$ and $$\Im(G(u+iv)) = \frac{2v}{(1-u)^2+v^2}$$

and in order to proof we are mapping onto $$\mathbb{H}$$ i would purely pay attention to the imaginary part, however, the author obviously uses the real part for his proof. What am i missing?

Thank you very much for any hint!

He is just using the fact that $$\Im (i(a+ib))=\Re (a+ib)$$ for any $$a,b \in \mathbb R$$. Take $$a+ib =\frac {1-w} {1+w}$$.