assignment7_soln

Contour integrals of f are independent of path we get

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Unformatted text preview: f (z ) dz + f (z ) dz = Γ Γ2 Γ1 f (z ) dz = 0 f (z ) dz − = −Γ2 Γ1 Section 4.4 9. Which of the following domains are simply connected? (a) The horizontal strip | Im z | < 1 Solution: This is simply connected. (b) The annulus 1 < |z | < 2 Solution: This is not simply connected. (c) the set of all point in the plane except those on the nonpositive x-axis. Solution: This is simply connected. 12. Given that D is a domain containing the closed contour Γ, that z0 is a point not in (z − z0 )−1 dz = 0, explain why D is not simply connected. D , and that Γ Solution: If D were simply connected the result would contradict Cauchy’s Integral Theorem. 3 13. Evaluate below. 1 dz along the three closed contours Γ1 , Γ2 , and Γ3 in the figure z2 + 1 Solution: We have 1 z 2 +1 Γ1 Γ2 Γ3 dz = i/2 z +i − i/2 . z −i Thus, i i 1 1 1 dz = dz − dz +1 2 Γ1 z + i 2 Γ1 z − i i i = (0) − (2πi) = π 2 2 1 i i 1 1 dz = dz − dz 2+1 z 2 Γ2 z + i 2 Γ2 z − i i i = (2πi) − (2πi) = 0 2 2 1 1 i i 1 dz = dz − dz 2+1 z 2 Γ3 z + i 2 Γ3 z − i i i = (2πi) − (0) = −π 2 2 z2 16. Show that if f is of the form Ak Ak−1 A1 + k −1 + · · · + + g (z ) (k ≥ 1) zk z z where g is analytic inside and on the circle |z | = 1, then f (z ) = f (z ) dz = 2πiA1 |z |=1 Solution: We have f (z ) dz = |z |=1 Ak dz + · · · + k |z |=1 z A1 dz + |z |=1 z g (z ) dz |z |=1 = 0 + · · · + 0 + 2πiA1 + 0 The first k − 1 integrals are 0 by path independence, the last integral is zero by Cauchy’s Integral Theorem, and the remaining integral is 2πi by the theorem from class. 4...
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This note was uploaded on 09/17/2013 for the course AMATH 332 taught by Professor R.moraru during the Spring '11 term at Waterloo.

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