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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:

enter image description here

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!

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

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  • $\begingroup$ Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again! $\endgroup$ – Zest Jan 12 at 23:52

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