Barbielly
幼苗
共回答了21个问题采纳率:95.2% 举报
调和函数的定义为∂^2(u)/∂x^2+∂^2(u)/∂y^2=0.
对u(x,y)=x^2+xy-y^2,
∂u/∂x=2x+y,所以∂^2(u)/∂x^2=2;
∂u/∂y=x-2y,所以∂^2(u)/∂y^2=-2.
所以∂^2(u)/∂x^2+∂^2(u)/∂y^2=0,u为调和函数
设f(z)=u+iv为解析函数,则由Cauchy-Riemann方程知
∂v/∂x=-∂u/∂y=-x+2y;
∂v/∂y=∂u/∂x=2x+y.
所以v=-x^2/2+2xy+y^2/2+C,C为任意常数.
所以
f(z)=u+iv
=x^2+xy-y^2+i(-x^2/2+2xy+y^2/2+C)
=(1-i/2)(x^2+2ixy-y^2)+iC
=(1-i/2)(x+iy)^2+iC
=(1-i/2)z^2+iC,
再将f(i)=-1+i代入可求得C=1/2,所以
f(z)=(1-i/2)z^2+i/2
1年前
10