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The radius of the largest (open) disc of centrum is therefore the distance between the z0 = −1 + i, in which the series is convergent towards Log z, √ point of expansion −1 + i and the negative real axis, thus 1 < 2. 7 Find the ﬁrst ﬁve terms of the power series expansion in z from z 0 = 0 for the following functions: (a) ez sin z (b) ez Log(1+z) , , (c) ez Log(1 + z). 2 Even if it is very easy to solve the task in MAPLE by using the command taylor, it is nevertheless a good exercise to try the more old-fashioned o-technique.

1 One can also set up another expression by computing all the coeﬃcients. If n = 8p, then a8p = 1 1 1 √ · √ = 4p+1 . com 39 Complex Functions Examples c-4 Analytic functions described as power series If n = 8p + 1, then a8p+1 = − 1 24p+1 . If n = 8p + 2, then a8p+2 = 1 24p+1 1 1 √ · √ = 4p+2 . 2 2 2 If n = 8p + 3, then a8p+3 = 0. If n = 8p + 4, then a8p+4 = 1 1 1 · √ · −√ 24p+2 2 2 =− 1 . 24p+3 =− 1 . 24p+4 If n = 8p + 5, then a8p+5 = −1 1 · (−1) = 4p+3 . 24p+3 2 If n = 8p + 6, then a8p+6 = 1 1 1 · √ · −√ 24p+3 2 2 If n = 8p + 7, then a8p+7 = 0.

N=0 It follows by the identity theorem that b0 = 1, b1 = −b0 = −1 and bn+2 = −bn+1 − 2bn , n ∈ N0 , so the two sequences (an ) and (bn ) fulﬁl the same diﬀerence equation. The solution of this is unique, so we conclude that bn = an , n ∈ N0 . It also follows from the above that the radius of 1 convergence is r = √ . 2 2) Then √ −1 + i 7 a= 4 and √ −1 − i 7 b= 4 are the roots of the polynomial. Since a · b = 1 1 1 , we have = 2b and = 2a. 2 a b 1 Then by a decomposition in the disc |z| < √ , 2 f (z) = = = 1 1 1 1 1 1 1 1 = = · · + · · 2(z − a)(z − b) 2 b−a z−a 2 a−b z−b 2z 2 + z + 1 ⎧ ⎫ 1 −1 1 1 −1 1 1 1 ⎨1 1 1 1 ⎬ + · · = · · − · · · 2 (b − a)a 1 − z 2 (a − b)b 1 − z 2 a − b ⎩a 1 − z b 1− z⎭ a b a b 1 1 · 2 a−b +∞ = n=0 2n · b 2b · n+1 1 1 − 2a · 1 − 2bz 1 − 2az −a b−a n+1 zn = +∞ = 1 a−b +∞ b n=0 (2b)n z n − a +∞ (2a)n z n n=0 an z n .