Chem Differential Eq HW Solutions Fall 2011 156

# Chem Differential Eq HW Solutions Fall 2011 156 - 156...

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156 Chapter 12 Green’s Functions and Conformal Mappings (b) By defnition oF the since, we have sin( ix )= e i ( ix ) - e - i ( ix ) 2 i = e - x - e x 2 i = i e x - e - x 2 = i sinh x. 29. (a) Note that, For z = x + iy ± =0 , x + x 2 + y 2 = z z · z = 1 z . We claim that this Function is not analytic at any z . We have u = x x 2 + y 2 ,v = y x 2 + y 2 ; so u x = y 2 - x 2 ( x 2 + y 2 ) 2 ,u y = - 2 xy ( x 2 + y 2 ) 2 x = - 2 xy ( x 2 + y 2 ) 2 y = x 2 - y 2 ( x 2 + y 2 ) 2 . Since the euqality u x = v y and u y = - v x imply that ( x, y )=(0 , 0). Hence f is not analytic at any z = x + . (b) Note that, For z = x + ± , x - x 2 + y 2 = z z · z = 1 z , and ths Function is anaytic For all z ± = 0. Using the Cauchy-Riemann equations, we have u = x x 2 + y 2 = - y x 2 + y 2 ; so u x = y 2 - x 2 ( x 2 + y 2 ) 2 y = - 2 xy ( x 2 + y 2 ) 2 x = 2 xy ( x 2 + y 2 ) 2 y = y 2 - x 2 ( x 2 + y 2 ) 2 . We have u x = v y and that u y = - v x . Hence the Cauchy-Riemann equations hold. Also, all the partial derivatives are continuous Functions oF ( x, y ) ± =(0 , 0). Hence by Theorem 1, f ( z 1 z is analytic For all z ± = 0 and f ± ( z u x + iv x = y 2 - x 2 ( x 2 + y 2 ) 2 + i 2 xy ( x 2 + y 2 ) 2 = ( y + ix ) 2 ( x 2 + y 2 ) 2 = ( y + ix ) 2 ( x 2 + y 2 ) 2 = [ i ( x - iy )]
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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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