By Beck M., Marchesi G., Pixton G.

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37 CHAPTER 4. INTEGRATION 38 (a) Let γ be the line segment from z = 0 to z = 1 + i. A parametrization of this curve is γ(t) = t + it, 0 ≤ t ≤ 1. We have γ (t) = 1 + i and f (γ(t)) = (t − it)2 , and hence 1 f= γ 1 (t − it)2 (1 + i) dt = (1 + i) 0 0 2 t2 − 2it2 − t2 dt = −2i(1 + i)/3 = (1 − i) . 3 (b) Let γ be the arc of the parabola y = x2 from z = 0 to z = 1 + i. A parametrization of this curve is γ(t) = t + it2 , 0 ≤ t ≤ 1. Now we have γ (t) = 1 + 2it and f (γ(t)) = t2 − t2 2 − i 2t · t2 = t2 − t4 − 2it3 , whence 1 1 t2 − t4 − 2it3 (1 + 2it) dt = f= γ 0 t2 + 3t4 − 2it5 dt = 0 1 1 14 i 1 + 3 − 2i = − .

16. Compute the real integral 2π 0 dθ 2 + sin θ by writing the sine function in terms of the exponential function and making the substitution z = eiθ to turn the real into a complex integral. 17. Show that F (z) = F (z) = arctan z? i 2 Log(z + i) − i 2 Log(z − i) is a primitive of 1 1+z 2 for Re(z) > 0. Is 18. Prove the following integration by parts statement. Let f and g be analytic in G, and suppose γ ⊂ G is a smooth curve from a to b. Then f g = f (γ(b))g(γ(b)) − f (γ(a))g(γ(a)) − γ f g. γ 19.

Then there is a unique M¨ obius transformation h satisfying h(z1 ) = w1 , h(z2 ) = w2 and h(z3 ) = w3 . Proof. Let h = g −1 ◦ f where f (z) = [z, z1 , z2 , z3 ] and g(w) = [w, w1 , w2 , w3 ]. 6. This theorem gives an explicit way to determine h from the points zj and wj but, in practice, it is often easier to determine h directly from the conditions f (zk ) = wk (by solving for a, b, c and d). 3 Exponential and Trigonometric Functions To define the complex exponential function, we once more borrow concepts from calculus, namely the real exponential function2 and the real sine and cosine, and—in addition—finally make sense of the notation eit = cos t + i sin t.