常数,则

1f(z)

为常数,故f(z)为常数.

16.解:(1)因为 u x2 xy y2,所以有 ux 2x y vy 2x y

v 2xy

y

2

2

c(x)

vx 2y c (x) uy 2y x

c (x) x c(x)

x

2

2

D

f(z) (x xy y) (2xy

i

22

y

2

2

x

2

2

D)i

12

由已知f(i)=-1+i -1+i=-1+ Di D

2

12)

f(z) (x xy y) i(2xy

22

y

2

2

x

2

2

(2)由C R条件,vy ux ex(xcosy ysiny) excoy,则 v (xexcosy exysiny excoy)dy xexsiny exsiny exysinydy xexsiny exycosy (x) 又因uy vx,故

esiny esiny eycosy (esiny xesiny ecosy (x))

x

x

x

x

x

x

即 (x) 0, (x) C,故

f(z) e(xcosy ysiny) i(xesiny eycosy C)

x

x

x

又因f(0) 0,故f(0) iC 0 C 0,所以 f(z) ex(xcosy ysiny) i(xexsiny exycosy) (3) 由C R条件, uy vx

2xy(x y)

2

2

2

,所以