分别由下列条件求解析函数f(z)=u+iv． (1)u=x2+xy—y2 f(i)=一1+i； (

正确答案：(1)由C．一R．方程v_y=u_x=2x+y则(2)由C．一R．方程v_y=u_x=e^x(xcos y—ysin y)+e^x．cos y则 v=∫(xe^x cos y—e^xysin y+e^xcos y)dy =xe^x sin y+e^xsin y—∫e^xysin ydy =xe^xsin y+e^xsin y—e^x(一ycos y+∫cos ydy) =xe^xsin y+e^xsin y+e^xcos y—e^xsin y+φ(x)=xe^xsin y+e^xycos y+φ(x) 又因u_y=一v_x故 一e^xsin y—e^xsin了一e^xy cos y=一(e^xsin y+xxe^xsin y+e^xcos y+'(x)) 即=C．故 f(z)=e^x(xcos y—ysin y)+i(xe^xsin y+e^xycos y+C) 又因f(0)=0故f(0)=iC=0→C=0故f(z)=e^x(xcos y—ysin y)+ie^x(xsin y+ycos y)．
(1)由C．一R．方程,vy=ux=2x+y,则(2)由C．一R．方程,vy=ux=ex(xcosy—ysiny)+ex．cosy,则v=∫(xexcosy—exysiny+excosy)dy=xexsiny+exsiny—∫exysinydy=xexsiny+exsiny—ex(一ycosy+∫cosydy)=xexsiny+exsiny+excosy—exsiny+φ(x)=xexsiny+exycosy+φ(x),又因uy=一vx,故一exsiny—exsin了一exycosy=一(exsiny+xxexsiny+excosy+'(x))即=C．故f(z)=ex(xcosy—ysiny)+i(xexsiny+exycosy+C)又因f(0)=0,故f(0)=iC=0→C=0,故f(z)=ex(xcosy—ysiny)+iex(xsiny+ycosy)．