06/22/2009, 02:24 PM
(06/22/2009, 12:37 AM)Tetratophile Wrote: and added "f has no other fixed points", because I thought other fixed points would mess things up.Well, but \( \exp \) *has* more fixed points. In every strip \( 2\pi i k < \Im(z) < 2\pi i (k+1) \) there is a fixed point of \( \exp \).
Quote:(3) We can repeat (4.2) as many times as needed to get to \( D_2 \).
No, we can not.
1. The curve \( f(\gamma) \) may get out of the strip of bijectivity of \( f \) what indeed happens for \( f=\exp \). (plot the curves!)
2. \( f(\gamma) \) may hit \( D_1 \), i.e. the extension of \( A_1 \) overlaps with the original domain. If \( \gamma \) meets \( L \) at an angle \( \alpha \) then \( f(\gamma) \) meets \( L \) at an angle \( \alpha+\arg(f'(L)) \). So the image curves rotate around L. This includes also the comment of Ben, that if \( \arg(f'(L))=0 \), i.e. if \( f'(L)) \) is real, then then the images do not rotate, and hence one would never reach \( D_2 \) if it hits \( L \) in a different angle.
