Originally Posted by

**topsquark** There is a slight problem here, but it doesn't affect the result.

$\displaystyle e^{iz} = e^{ir \, cos \theta} e^{i^2r \, sin \theta} = e^{-r sin \theta}e^{ir \, cos \theta}$

So the real part of e^(iz) is e^(-r sin(theta)), not the cosine version. Additionally this means that |e^(iz)| = |e^(-r sin(theta))| -> 0 as r -> infinity, so it is the |e^(-iz)| term that dominates as r -> infinity. But as this term also goes to infinity, we are merely switching the dominant term and the proof goes through otherwise unaffected.

-Dan