+ic,为使f(i)=0,当x=0,y=1时,f(i)=0,有f(0)=-i+ic=0∴c=1∴f(z)=Z3+i4、设f(z)=u+iv在区域G内为解析函数,且满足u=2(x-1)y,f(2)=-i,试求f(z)。解:依C-R条件有Vy=ux=2y∴V==y2+(x)∴Vx=∴(x)=V=y2-x2+2x+c(c为常数)∴f(z)=2(x-1)y+i(y2-x2+2x+c)为使f(z)=-i,当x=2y=0时,f(2)=ci=-i∴c=-1∴f(z)=2(x-1)y+i(y2-x2+2x-1)=-(z-1)2i四、证明题1、试在复平面讨论f(z)=iz的解析性。解:令f(z)=u+ivz=x+iy则iz=i(x+iy)=-y+ix∴u=-yv=x于是ux=0uy=-1Vx=1Vy=0∵ux、uy、vx在复平面内处处连接又Ux=VyUy=-Vx。∴f(z)=iz在复平面解析。2、试证:若函数f(z)在区域G内为解析函数,且满足条件(z)=0,z∈G,则f(z)在G内为常数。证:设f(z)=u+iv,z=x+iy,z∈G∵f(z)在G内解析,Ux=Vy,Uy=-Vx又(z)=0,(z)=Ux+iVxUx=0Vx=0Uy=-Vx=0Ux=Vy=0U为实常数C1,V也为实常数C2,f(z)=C1+iC2=Z0f(z)在G内为常数。复变函数课程作业参考解答2第3章初等函数一、单项选择题1.z=(A)是根式函数的支点.(A)0(B)1(C)(D)i2.z=(D)是函数的支点.(A)i(B)2i(C)-1(D)03.ei=(B).(A)e-1+e(B)cos1+isin1(C)sin1(D)cos14.sin1=(A)(A)(B)(C)(D)二、填空题1.cosi=2.=e(cos1+isin1)3.lni=4.ln(1+i)=k为整数.三、计算题1.设z=x+iy,计算.解:∴∴==
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