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By Karl-Heinz Fieseler

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**Additional resources for Complex Analysis [Lecture notes]**

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Furthermore a ∈ Vb =⇒ Dr (a) ⊂ Vb with r = dist(a, K). e. Vb = C \ K. Indeed let c ∈ Dr (a). Then, for |z − a| > |c − a| we have ∞ 1 (c − a)ν =− ∈ La (K) ⊂ Lb (K), c−z (z − a)ν+1 ν=0 since the right hand side converges uniformly on C \ D (0) for any |c − a|. 3 (Theorem of Mittag-Leffler). Let G ⊂ C be a domain. Given a sequence (an )n∈N ⊂ G without accumulation point in G and polynomials rn cnk pn (z) = (z − an )k k=1 62 1 without constant term, there is a meromorphic function having the in z−a n points an as its poles with pn (z) as the principal part of its Laurent series around the point an ∈ G.

H ∈ O(C \ S) is the derivative of a function H ∈ O(C \ S), 3. ca = Resa (f ) for a ∈ S. is a rational function and S denotes the set of zeros If G = C and f (z) = p(z) q(z) of the polynomial q(z), then g(z) is the polynomial part of the partial fraction decomposition of f and h(z) the sum of all singular terms of multiplicity > 1. Given the decomposition we obtain the residue formula as follows: For sufficiently small ε > 0 integration over Dε (a) yields ca = Resa (f ), while integration over λ gives the desired formula: We have λ g(z)dz = 0, since λ is nullhomologous in G and λ h(z)dz = 0 because of hdz = dH.

A removable singularity of f if there is a function f ∈ O(Ga ) extending f. 2. a pole of f if a is not a removable singularity, but there is some n ∈ N>0 , auch that the function g(z) = (z − a)n f (z) has at a a removable singularity. 3. an essential singularity of f otherwise. 16. Let f ∈ O(G) be a holomorphic function on the domain G and a ∈ C an isolated boundary point of G with Dr (a)∗ ⊂ G, write f (z) = ∞ n n=−∞ an (z − a) . Then the point a is 1. a removable singularity iff f |D for n < 0, (a)∗ is bounded for some > 0 iff an = 0 2.