hw5 - Mathematics 185 \u2013 Intro to Complex Analysis Fall 2009 \u2013 M Christ Solutions to Selecta from Problem Set 5 2.2.1 Prove that C{0 is simply

# hw5 - Mathematics 185 – Intro to Complex Analysis Fall...

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Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 5 2.2.1 Prove that C \ { 0 } is simply connected. Solution. According to the homotopy form of Cauchy’s Theorem, Theorem 2.3.14, if an open set G is simply connected, then for any closed contour γ in G , and for any function f analytic in G , R γ f = 0. But we have already calculated that R Γ dz z = 2 πi , where Γ is the unit circle θ 7→ e , θ [0 , 2 π ]. Γ is a closed contour in G = C \ { 0 } , and f ( z ) = z - 1 is certainly analytic in G , so C \ { 0 } cannot be analytic. 2.2.7(a) Evaluate the following integral without doing any explicit computations: R γ z - 1 dz , where γ is the ellipse γ ( t ) = cos( t ) + 2 i sin( t ) for t [0 , 2 π ]. Solution. The integral equals 2 πi , because (i) f ( z ) = z - 1 is analytic in G = C \{ 0 } . (ii) This ellipse is homotopic in G to the circle ˜ γ ( t ) = e it = cos( t ) + i sin( t ); one homotopy is h ( s, t ) = cos( t ) + (1 + s ) i sin( t ). (iii) R ˜ γ z - 1 dz = 2 πi . 2.2.7(c) Evaluate without explicit computation: R γ z - 1 e z dz , where γ ( t ) = 2 + e it , t [0 , 2 π ].

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