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2 Winter 2000

# 2 Winter 2000 - Solutions to second midterm 132/1 Winter...

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Problem 1(a) (5 points): Let gamma_1 be the curve gamma_1(theta) = exp(i theta), 0 <= theta <= 3 pi . Compute the integral of z dz along gamma_1 , where z is the conjugate of z . Answer: Making the change of variables z = exp(i theta) dz = i exp(i theta) d theta and observing that z = exp(- i theta) the integral becomes that of exp(-i theta) i exp(i theta) d theta from 0 to 3pi. The exponentials cancel, so you are just integrating i dtheta from 0 to 3 pi, which is 3 pi i. Some common errors: Ignoring the conjugation, and writing z as exp(i theta) instead of exp(- i theta) . (Or ignoring the z term altogether). Using the fundamental theorem of calculus, and trying to press z^2/2 or z ^2/2 into service as an anti-derivative for z . (Actually z does not have an anti-derivative, because it is not analytic, and only analytic functions have anti-derivatives). Other comments: It is possible to "guess" the answer as 3 pi i, based on reasoning by analogy with similar examples. However, this type of reasoning can be dangerous. For instance, suppose we are integrating z on the curve z = 2 exp(i theta), 0 <= theta <= 2 pi. Can you guess the answer by inspection? Make a guess, and then check your answer. Problem 1(b) (10 points): Let gamma_2 be the line segment from -1 to -1-i. Compute the integral of dz/z on gamma_2. Answer: We shall apply the fundamental theorem of calculus. We want an anti-derivative of dz/z which is analytic on gamma_2. The principal branch Log(z) won't work, because it's not analytic on the negative real line, and gamma_2 just touches this line. However, a branch such as Log_(0,2 pi] (z) is analytic on gamma_2 (it is only non-analytic at 0 and the positive real axis, which is not a problem). So by the Fundamental theorem of calculus the answer is Log_(0,2 pi](-1-i) - Log_(0,2 pi](-1).

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