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By Karl-Heinz Fieseler

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12. Given a function f ∈ O(A ,r (a)) there are unique functions f+ ∈ O(Dr (0)) and f− ∈ O(D −1 (0)) (where ”0−1 = ∞” and D∞ (0) = C) with f− (0) = 0, such that f (z) = f+ (z − a) + f− 1 z−a . In particular there are complex numbers an , n ∈ Z, such that ∞ ∞ n an (z − a) := f (z) = n=−∞ ∞ a− (z − a)− . n an (z − a) + n=0 53 =1 A series of the above type is called a Laurent series, the first part is called the power series part of the Laurent series, the second one its principal part. Indeed 1 f (z)dz an = , < s < r.

We have a0 = 0, and the function h ∈ O(G) is continuous, hence h(z) = 0 on some disc Dr (z0 ) and thus f (z) = (z − z0 )n h(z) = 0 for z ∈ Dr (z0 )∗ . 10 (Identity theorem). Let A ⊂ G be a subset with a point z0 of accumulation within G and f, g ∈ O(G). If f |A = g|A , then f = g. Proof. The zeros of the function f − g ∈ O(G) are not isolated, since by continuity, f (z0 ) − g(z0 ) = 0. As a consequence of the previous lemma we have f − g ≡ 0. 11. Let f : G −→ C be a holomorphic function of multiplicity n at z0 ∈ G.

In particular we obtain f |K ∈ Lb (K). Now choose a closed disc D := Dr (0) ⊃ K and b ∈ D. Denote q(z) a 1 polynomial in z−b with ||f − q||K < ε/2 and take p(z) as the n-th Taylor polynomial of q around 0. Then ||q − p||D < ε/2 for n 0 and thus ||f − p||K < ε. 1 with a ∈ It remains to show that all degree one Laurent monomials a−z C \ K belong to Lb (K). We consider the set Vb := a ∈ C \ K; 1 ∈ Lb (K) . a−z Since b ∈ Vb , we have Vb = ∅. Furthermore a ∈ Vb =⇒ Dr (a) ⊂ Vb with r = dist(a, K). e.

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