(07/06/2009, 08:56 AM)bo198214 Wrote: One needs to show that then there is always a pair of points \( c_1=\gamma(t_1) \) and \( c_2=\gamma(t_2) \) with equal real part and with \( \Im(c_2)-\Im(c_1)=2\pi \).
This is equivalent to that \( \gamma \) and \( \gamma+2\pi i \) intersect.
If \( \gamma \) only extend to the right, i.e. \( \Re(\gamma(t))>\Re(a)\forall t\in (0,1) \), then this is a consequence of the Jordan curve theorem. We have \( \Im(a) < \Im(a+2\pi i) < \Im(b) < \Im(b+2\pi i) \). The closed Jordan curve \( [a,b]\cup \gamma \) has the point \( \gamma(0+\epsilon)+2\pi i\approx a+2\pi i \) in its interior and the point \( \gamma(1-\epsilon)+2\pi i\approx b+2\pi i \) in its exterior. Hence there must be an intersection of \( \gamma+2\pi i \) and \( \gamma \) (as \( \gamma+2\pi i \) does not pass [a,b] for \( t\in (0,1) \).)
