which, again, we can solve using a Cauchy integral and (31). The problem of finding K± is more delicate. At first sight, we could take the logarithm of (33), giving logK+− logK− = logG. This looks similar to (31), but it usually happens that logG(t) is not continuous for all t ∈ C , which means that we cannot use (30). However, this difficulty can be overcome. The problem of finding K± such tha...