五(7分).如果f(x)=x2+g(y)+iv(x,y)是解析函数,且f()=f'(0)=0,求实 函数g(y),v(x,y)及f(x).提示:根据x2+9(y)是调和函数求g(y) 解因为u(x,y)=x2+g(y)是调和函数,所以 zx+uy=2+g"(y)=0 所以g"(y)=-2,得9(y)=-y2+C1y+C2,(3分) 根据C-R方程,有 2x,所以 v(a, y) +h(a)=2 ry+h(a), 又因为tx=-uy=2y-C1,于是有 2y+h(a)=2y-C 所以h(x)=-C1,得h(x)=-C1x+C3于是,有 v(a, y) Clr+C3 所以 f(z) +C1y+C2+(2xy-C1x+C3)(3分 根据f(0)=0,得:C2=C3=0.根据f(0)=0,得:C1=0.所以 f()=x2-y2+2xryi.(1分) 六(6分)已知f(x)在0<|z<1内解析,且 lim zf(2)=1 (1)证明:z=0是f(x)的一级极点; (2)求Res(2),0.提示:考虑xf(2)在0<|2<1内的洛朗级数 证因为zf(z)在0<|z<1内解析,并且1imxf(x)=1,所以f()在该圆环 域内的洛朗级数是 zf(2)=1+c1z+c2+c323+ (3分) 于是 f(x)=-+c1+c2z+ (1分) 所以z=0是f(z)的一级极点(1分),且Resf(z),]=1(1分)6  (7 +). t7 f(z) = x 2 + g(y) + iv(x, y) ~H9
￾n f(0) = f 0 (0) = 0, o{ 9
g(y), v(x, y) A f(z). [}2J x 2 + g(y) ~#;9
o g(y)] B " u(x, y) = x 2 + g(y) ~#;9
￾ uxx + uyy = 2 + g 00(y) = 0,  g 00(y) = −2, g(y) = −y 2 + C1y + C2. (3 +) 2J C-R *￾& vy = ux = 2x,  v(x, y) = Z 2xdy + h(x) = 2xy + h(x), '" vx = −uy = 2y − C1, (~& 2y + h 0 (x) = 2y − C1,  h 0 (x) = −C1, h(x) = −C1x + C3. (~￾& v(x, y) = 2xy − C1x + C3.  f(z) = x 2 − y 2 + C1y + C2 + (2xy − C1x + C3)i (3 +) 2J f(0) = 0, C2 = C3 = 0. 2J f 0 (0) = 0, C1 = 0.  f(z) = x 2 − y 2 + 2xyi. (1 +) \ (6 +). 3 f(z) , 0 < |z| < 1 fH￾n lim z→0 zf(z) = 1. (1) 2d z = 0 ~ f(z) !C@" (2) o Res[f(z), 0]. [}L^ zf(z) , 0 < |z| < 1 f!_SC
] H " zf(z) , 0 < |z| < 1 fH￾n lim z→0 zf(z) = 1,  zf(z) ,/+> )f!_SC
~ zf(z) = 1 + c1z + c2z 2 + c3z 3 + · · · , (3 +) (~ f(z) = 1 z + c1 + c2z + c3z 2 + · · · , (1 +)  z = 0 ~ f(z) !C@" (1 +), n Res[f(z), 0] = 1(1 +).