(07/06/2009, 08:56 AM)bo198214 Wrote: One needs to show that then there is always a pair of points and with equal real part and with .

This is equivalent to that and intersect.

If only extend to the right, i.e. , then this is a consequence of the Jordan curve theorem. We have . The closed Jordan curve has the point in its interior and the point in its exterior. Hence there must be an intersection of and (as does not pass [a,b] for .)