156Chapter 12Green’s Functions and Conformal Mappings(b) By defnition oF the since, we havesin(ix)=ei(ix)-e-i(ix)2i=e-x-ex2i=iex-e-x2=isinhx.29.(a) Note that, Forz=x+iy±=0,x+x2+y2=zz·z=1z.We claim that this Function is not analytic at anyz. We haveu=xx2+y2,v=yx2+y2;soux=y2-x2(x2+y2)2,uy=-2xy(x2+y2)2x=-2xy(x2+y2)2y=x2-y2(x2+y2)2.Since the euqalityux=vyanduy=-vximply that (x, y)=(0,0). Hencefisnot analytic at anyz=x+.(b) Note that, Forz=x+±,x-x2+y2=zz·z=1z,and ths Function is anaytic For allz±= 0. Using the Cauchy-Riemann equations, wehaveu=xx2+y2=-yx2+y2;soux=y2-x2(x2+y2)2y=-2xy(x2+y2)2x=2xy(x2+y2)2y=y2-x2(x2+y2)2.We haveux=vyand thatuy=-vx. Hence the Cauchy-Riemann equations hold.Also, all the partial derivatives are continuous Functions oF (x, y)±=(0,0). Henceby Theorem 1,f(z1zis analytic For allz±= 0 andf±(zux+ivx=y2-x2(x2+y2)2+i2xy(x2+y2)2=(y+ix)2(x2+y2)2=(y+ix)2(x2+y2)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.