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When this equation is differentiated we get (1 − w) d3 w dw d2 w dw d2 w − · − 2(2 − w) · + dz 3 dz dz 2 dz dz 2 dw dz 3 = 0, hence by insertion of the previous results, d3 w dz 3 z=0 = 1 · 2 + 2 · 2 · 1 · 2 − 1 = 9. Finally, by insertion into Taylor’s formula we obtain in a neighbourbood of z 0 = 0 that w(z) = 0 + 1 2 9 3 z + x2 + z 3 · · · = z + z 2 + z 3 + · · · . 1! 2! 3! 16 Lad M ∈ R. Prove that if f (z) is analytic in C, and Re(f (z)) ≤ M for every z, then f (z) is constant. Hint: Apply Liouville’s theorem on exp(f (z)).

4n na na n=1 n=1 1 it follows that the w-radius of convergence is na Putting cn = = lim w 1 n→+∞ n |cn | = lim na/n = lim exp n→+∞ n→+∞ a ln n = exp(0) = 1. e. e. for |z| > 2. Then it follows from the definition that the z-radius of convergence is z = 2. 16 Given a series convergence for each of the series +inf ty (a) +∞ cn z 2n , (b) n=0 +∞ n n=0 cn z nn cn z n , n=0 of radius of convergence +∞ (c) ∈ R+ . Find the radius of (2n − 1) cn z n . n=0 It follows from the assumption that lim sup where 0 < n |cn | = 1 , < +∞.

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. Summing up we get for |z − 1| < √ 2, +∞ 1 1 = n 1 + z2 2·16 n=0 1−(z−1)+ (z−1)4 (z−1)5 (z−1)6 (z−1)2 − + − 2 4 4 8 (z−1)8n . 4 Find the Taylor series from z0 = 1 for each of the following functions, and indicate the radius of convergence of the series: (a) 1 , z−2 (b) 1 . z(z − 2) (a) By a straightforward computation, +∞ 1 1 (z − 1)n =− =− z−2 1 − (z − 1) n=0 for |z − 1| < 1.

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