final_solutions_185

# final_solutions_185 - Complex Analysis Math 185A Winter...

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Complex Analysis Math 185A, Winter 2010 Final: Solutions 1. [25 pts] The Jacobian of two real-valued functions u ( x,y ), v ( x,y ) of ( x,y ) is deﬁned by the determinant J = ( u,v ) ( x,y ) = ± ± ± ± u x u y v x v y ± ± ± ± . If f ( z ) = u ( x,y ) + iv ( x,y ) is an analytic function of z = x + iy , prove that J ( x,y ) = | f 0 ( z ) | 2 . Solution. Since f is analytic, f 0 = u x + iv x = v y - iu y and u , v satisfy the Cauchy-Riemann equations u x = v y , v x = - u y . It follows that J = ± ± ± ± u x u y v x v y ± ± ± ± = u x v y - u y v x = u 2 x + v 2 x = | f 0 | 2 . Remark. Note that the Jacobian is strictly positive when f 0 6 = 0, corre- sponding to the fact that a conformal map f is locally one-to-one. Also, J is always nonnegative which corresponds to the fact that analytic functions preserve orientations. 1

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2. [25 pts] Deﬁne f ( z ) = z + 1 z . (a) Find the image of the unit circle | z | = 1 under f . (b) On what open sets Ω C is f : Ω C a conformal map? Solution. (a) If | z | = 1 then z = e for some θ [0 , 2 π ). Then z + 1 z = e + e - = 2 cos θ. So f maps the unit circle to the interval [ - 2 , 2] on the real axis. (b) The function f is analytic except at z = 0. Its derivative is then f 0 ( z ) = 1 - 1 z 2 which is nonzero unless z = ± 1. Therefore f is conformal on any open set Ω C that does not contain any of the points {- 1 , 0 , 1 } . Remark. Note that f has a simple zero at z = ± 1, so it doubles the angle between curves at those points, where it ‘ﬂattens’ the circle | z | = 1 to the interval [ - 2 , 2]. This mapping is used in ﬂuid mechanics to obtain the potential ﬂow of an ideal ﬂuid past a cylinder of radius one, which f maps to uniform ﬂow past a ﬂat plate of length four. 2
3. [25 pts] Let γ : [0 ] C with γ ( t ) = 2 e it be the positively oriented semicircle in the upper half plane with center the origin and radius 2. Prove that ± ± ± ± Z γ e z z 2 + 1 dz ± ± ± ± 2 πe 2 3 . (Do not try to evaluate the integral exactly.) Solution. By the basic estimate for the modulus of a contour integral,

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final_solutions_185 - Complex Analysis Math 185A Winter...

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