“…If β is periodic, then there exists N ∈ N, N ≥ 2 such that β N = 1 and β k = 1 for k < N. It follows that for all f ∈ Hol(D),T N (f )(z) = m N (z)f (z), m N (z) = Hence, by [4, Proposition 3.2], we have σ(T ) = {λ ∈ C : λ N ∈ m N (D)}. r If β is aperiodic, then by [4, Corollary 2.2], {β n B(0) : n ∈ N 0 } ⊂ σ(T ).Moreover, using [4, Proposition 7.2], we have σ(T ) ⊂ D(0, M), with M defined by Re it /r) dt .Note that if m(z) = B(z/r), we can writem(z) = z N m(z), with m(z) = e iθ r N K j=1 z − rα j r − α j z , with α 1 , • • • , α K = 0.Since m is well defined in D, the equation (7.2) of[4] is also valid for M R with R → 1, that is for M 1 = M. Hence, .Therefore, if B has d zeroes (counting the multiplicities), then {β n B(0) : n ∈ N 0 } ⊂ σ(T ) ⊂ D(0, r −d ).…”