“…Notice that Γ < ∞ (and hence ∆ > 0), since α is bounded away from ∂D. Moreover, we also have Γ ≥ 1 (and hence ∆ ≤ η) since det(S(α, z 0 )) = z 0 for all α ∈ D. If n k is even, one may apply the arguments from [30] without modification to deduce that max( Φ(2n k ) , Φ(n k ) ) = max( Φ(2n k ) , Φ(n k ) ) ≥ η ≥ ∆, since (1.8) implies that |tr(Z(j, 0; z))| = |tr(T (j, 0; z))| whenever j is even and z ∈ ∂D. Now, suppose n k is odd.…”