Chem Differential Eq HW Solutions Fall 2011 160

Chem Differential Eq HW Solutions Fall 2011 160 - e ,...

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160 Chapter 12 Green’s Functions and Conformal Mappings Solutions to Exercises 12.6 1. (a) f ( z )=1 /z is analytic for all z ± = 0, by Theorem 2, Section 12.5, since it is the quotient of two analytic functions. U ( u, v )= uv is harmonic since U uu =0 , U vv = 0, so U uu + U vv =0 . (b) We have (this was done several times before) f ( z )= x - iy x 2 + y 2 = x x 2 + y 2 - i y x 2 + y 2 . So Re ( f )= u ( x, y )= x x 2 + y 2 and Im( f )= v ( x, y )= - y x 2 + y 2 . (c) We have φ ( x, y )= U f ( z )= U ( u ( x, y ) ,v ( x, y ) ) = U ± x x 2 + y 2 , - y x 2 + y 2 ² = - xy ( x 2 + y 2 ) 2 . You can verify directly that φ ( x, y ) is harmonic for all ( x, y ) ± =(0 , 0) or, better yet, you can apply Theorem 1. 5. (a) If z is in S , then z = a + i,y where b y c .So f ( z )= e z = e a + iy = e a e iy . The complex number w = e a e iy has modulus e a and argument y .A s y varies from b to c , the point w = e a e iy traces a circular arc of radius e a , bounded by the two rays at angles b and c . (b) According to (a), the image of { z =1+ iy :0 y π/ 2 } by the mapping e
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Unformatted text preview: e , bounded by the two rays at angles 0 and / 2. It is thus a quarter of a circle of radius e (see Fgure). Similarly, the image of { z = 1 + iy : 0 y } by the mapping e z is the circular arc with radius e , bounded by the two rays at angles 0 and . It is thus a semi-circle of radius e , centered at the origin (see Fgure). The most basic step is to define the complex variable z=x+iy, where x and y real. This is done as follows: Clear x, y, z, f AlgebraReIm x : Im x y : Im y z x I y You can now define any function of z and take its real and imaginary. For example: f z_ E^z Re f z Im f z x y x Cos y x Sin y...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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