12/28/2008, 11:14 AM
Thank you for being so patient with me, Dmitrii. Now it finally came through that if sexp maps \( A \) biholomorphically to \( C \) then it can also map \( B \) biholomorphically to \( C \) though \( A\cap B=\emptyset \) may be.
I now also see that the image of slog has a width of roughly \( \frac{2\pi}{\arg(L)} \) because an increment by 1 causes the image of sexp to revolve \( \arg(L)=\Im(L) \) around the fixed point (near the fixed point). And after \( 2\pi \) revolution it reaches the cut.
I now also see that the image of slog has a width of roughly \( \frac{2\pi}{\arg(L)} \) because an increment by 1 causes the image of sexp to revolve \( \arg(L)=\Im(L) \) around the fixed point (near the fixed point). And after \( 2\pi \) revolution it reaches the cut.