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November 24, 2010 Lecturer Dmitri Zaitsev Michaelmas Term 2010 Course 2325 2010 Complex Analysis I Sheet 3 Due: at the end of the lecture Exercise 1 (i) Show that (ez )0 = ez . (Hint. Differentiate in the direction of the x-axis.) Solution. Since f 0 (z0 ) = lim z→z0 = f (z0 + h) − f (z0 ) f (z) − f (z0 ) = lim = h→0 z − z0 h f (z0 + h) − f (z0 ) = ∂x f (z0 ) h∈IR, h→0 h lim holds for any holomorphic f (z), we have (ez )0 = ∂x ex (cosy + isiny) = ex (cosy + isiny) = ez . Exercise 2 (iii) Use (ii) to show that the conclusion of Cauchy’s theorem does not hold for f (z) = z̄. Solution. It follows from (ii) that Z z̄ dz 6= γ1 Z z̄ dz γ2 whenever γj corresponds to some yj with y1 6= y2 . Form a closed path C by following γ1 from −1 to 1 and then γ2 with reversed orientation from 1 to −1. Then Z z̄ dz = C Z z̄ dz − γ1 Z z̄ dz 6= 0. γ2 Thus we have a closed path for which the conclusion of Cauchy’s theorem does not hold with f (z) = z̄. Exercise 3 (i) Calculate R γ f (z) dz, where f (z) = Solution. Z γ 1 dz = z Z 1 z and γ(t) = eit , 0 ≤ t ≤ 2π, is the unit circle. 2π 0 ieit dz = 2πi. eit (ii) Use (i) to show that f (z) does not have an antiderivative in its domain of definition. Solution. It f had antiderivative, its integral would be zero over any closed path, which is not the case. (iii) Use Exercise 1 (ii) to give an example of a domain Ω, where f does have an antiderivative. Solution. Take Ω = C \ {0}, the domain of definition of f (z) = 1z , then (ii) shows that f does not have antiderivative in Ω. (iv) Does f (z) = 1 zn have an antiderivative, where n ≥ 2 is an integer? Solution. Yes, take F (z) = Justify your answer. 1 (−n+1)z n−1 .