解析f(z)=u+iv在复平面上为解析函数v=3x^2y-y^3+1,且知f(i)=0,求f(z).

Question

解析f(z)=u+iv在复平面上为解析函数v=3x^2y-y^3+1,且知f(i)=0,求f(z).

Answer 1

Vx'=6xy=-U'y, 因此U=-3xy^2+C1(x)
Vy'=3x^2-3y^2=U'x, 因此U=x^3-3xy^2+C2(y)
故有：C1(x)=x^3+C2(y)
因此C1(x)=x^3+C,C2(y)=C
U=-3xy^2+x^3+C
f(z)=(-3xy^2+x^3+C)+i(3x^2y-y^3+1)
由f(i)=C=0,
因此 f(z)=(-3xy^2+x^3)+i(3x^2y-y^3+1)