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MA 513 HW 5

# MA 513 HW 5 - MA 513 p 170#5 7 10 p 178 1 2 9 Homework 5...

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Unformatted text preview: MA 513 p. 170: #5, 7, 10; p. 178: # 1, 2, 9. Homework 5 Solutions p. 170: #5: C f ′ (z ) dz = 2πif ′ (z0 ) = z − z0 C f (z ) dz . (z − z0 )2 #7: so C eaz dz = 2πieaz z π −π π −π = 2πi, by the Cauchy Integral Formula. Now z = cos θ + i sin θ, z =0 C eaz dz = z = exp(a(cos θ + i sin θ)) i dθ = π −π exp(a cos θ) exp(ia sin θ)) i dθ π −π exp(a cos θ) (cos(a sin θ) + i sin(a sin θ)) i dθ = 2 exp(a(cos θ)) (cos(a sin θ) i dθ, using evenness and oddness of exp(a cos θ) cos(a sin θ), exp(a cos θ) sin(a sin θ) respectively. The answer follows by comparing the two results and dividing by 2i. #10: From the Cauchy Integral Formula, we have, for any R > |z |, |f ′′ (z )| = 1 πi AR f (s ) dz ≤ 2R → 0, 3 (R − | z | )3 | s | = R (s − z ) as R → ∞. Hence, f ′′ (z ) = 0. Therefore, f (z ) = a1 z + b1 for some complex constants a1 , b1 . But |f (0)| = |b1 | = 0, since |f (z )| ≤ A|z | for all z. p. 178: #1: Let g (z ) = exp[f (z )], as suggested. Then |g (z )| = exp(u(x, y )) ≤ exp(u0 ). Since f is entire, so is g. Hence, by Liouville’s theorem, g (z ) is constant, so |g (z )| = exp(u(x, y )) is constant. Thus, u(x, y ) is constant. #2: Discussed in class. #9: (a) Either multiply out, or use induction. (b) Note that P (z0 ) = 0, so n P (z ) = P (z ) − P (z0 ) = a1 (z − z0 ) + k =2 k ak (z k − z0 ). The result now follows from part (a). ...
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