Solutions for assignment 5

# Solutions for assignment 5 - Section 4.4 Question 10: (a)...

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Unformatted text preview: Section 4.4 Question 10: (a) The numerator and denominator are analytic, and the denominator is zero exactly when z is +5i or -5i, so the domain of analyticity is C - {5i, -5i}. Thus the function is analytic on and inside the contour |z| = 2, and Cauchy's theorem applies. (Note that the domain of analyticity is not simply connected, but we could make a smaller domain, e.g. {|z| &lt; 3} which is simply connected and still contains the contour. (c) Similar reasoning to (a), except that the two bad points are 3+i and 3-i. (e) sec(z/2) = 1 / cos(z/2); both the numerator and denominator are analytic, and the denominator is zero only when z = +-pi, +-3pi, +-5pi, ... (as can be seen by e.g. writing cos(z/2) = (exp(iz/2) + exp(-iz/2))/2 and solving). So the domain of analyticity is all the points which are not odd multiples of pi, so the function is analytic on and inside the contour. Question 13: Although this question precedes the Cauchy integral formula, the CIF is by far the best way to compute these integrals. (I don't know why the question is placed where it is). For (a), apply CIF with f(z) =compute these integrals....
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## Solutions for assignment 5 - Section 4.4 Question 10: (a)...

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