By Beck M., Marchesi G., Pixton G.
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Extra resources for A first course in complex analysis
1. As our first example of the application of this definition we will compute the integral of the function f (z) = z 2 = x2 − y 2 − i(2xy) over several curves from the point z = 0 to the point z = 1 + i. 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.
Because e is a positive real number and hence Arg e = 0, we obtain ez = exp(z Log e) = exp (z (ln |e| + i Arg e)) = exp (z ln e) = exp (z) . A word of caution: this only works out this nicely because we carefully defined ab for complex numbers. Different definitions might lead to different outcomes of ez versus exp z! 3 Named after Leonard Euler (1707–1783). html. CHAPTER 3. EXAMPLES OF FUNCTIONS 33 Exercises 1. Show that if f (z) = az+b cz+d is a M¨ obius transformation then f −1 (z) = dz−b −cz+a .
B) Show that the image of a ray starting at the origin is a ray starting at the origin. (c) Let T be the figure formed by the horizontal segment from 0 to 2, the circular arc from 2 to 2i, and then the vertical segment from 2i to 0. Draw T and f (T ). (d) Is the right angle at the origin in part (c) preserved? Is something wrong here? ) 34. As in the previous problem, let f (z) = z 2 . Let Q be the square with vertices at 0, 2, 2 + 2i and 2i. Draw f (Q) and identify the types of image curves corresponding to the segments from 2 to 2 + 2i and from 2 + 2i to 2i.
A first course in complex analysis by Beck M., Marchesi G., Pixton G.