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I've got the next question from Conway's textbook in complex analysis vol I.

In page 161-162 he defines $g$ there, here's an excerpt: (page 162)


I don't understand why $|g(z)|<1$, the notation is that D is the open unit disk.

Your help will be much appreicated.

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The link you provided does not lead where I think you meant it to –  Bey Oct 16 '12 at 11:29
press on the link "The Riemann mapping Theorem", in chapter VII, then scroll down to page 162. Did I really need to tell you that? which you could have done by yourself, by seeing that I mentioned the number of the page I am refering to... –  MathematicalPhysicist Oct 16 '12 at 11:57
Only about 50 pages of the book are shown, and these do not include pp. 161-162. –  Per Manne Oct 16 '12 at 12:12
It does include page 162, and he defines this function there. –  MathematicalPhysicist Oct 16 '12 at 12:22
I don't think the available pages shown on Google Books are the same for every user. For me it does not show the page either. So please try to copy the relevant part of the books into your question. –  Lukas Geyer Oct 16 '12 at 15:28

1 Answer 1

up vote 1 down vote accepted

In the cited section of Conway an injective holomorphic function $h\colon G\to D$ has been constructed, where $G\subset\mathbb{C}$ is a domain. For a point $a\in G$, one then defines a function $g\colon G\to D$ by $$ g(z) = \frac{|h'(a)|}{h'(a)}\frac{h(z) - h(a)}{1 - \overline{h(a)}h(z)}.$$ My understanding of the question is: why is it that $g$ maps into $D$?

The answer to this question is that the function $$\varphi(w) := \frac{|h'(a)|}{h'(a)}\frac{w - h(a)}{1 - \overline{h(a)}w}$$ is an automorphism of $D$ (see p. 131 in Conway). The function $g$ is the composition $\varphi\circ h$. Since $h\colon G\to D$ and $\varphi\colon D\to D$, the composition $g$ is from $G\to D$.

Also, it would be more helpful for us and for future users with the same question if you wrote the question explicitly instead of providing a (fickle) link. Hope this helps.

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Thanks, sorry for my attitude I am just in a bad mood. –  MathematicalPhysicist Oct 16 '12 at 18:17

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