“…To be speci®c, let D n h n H and D hH, where H fz: Im z > 0g. Then D is a component of ker n31 D n 1 n1 int 1 mn D m and h À1 n 3 h À1 pointwise in D. We shall produce a contradiction by demonstrating that the domain D is conformally homogeneous, and hence, in view of Theorem 3 in [7] (see also [13]), that ¶D is a circle in C. Since the points 0, 1 and 1 lie on ¶D, ¶D R È f1g. Thus hI, a subarc of ¶D that has endpoints 0 and 1 but does not contain 1, must coincide with I.…”