如题所述
(1)
∂v/∂x=2y+3
∂u/∂x=∂v/∂y=2x (柯西黎曼方程)
f‘(z)=∂u/∂x+i(∂v/∂x)=2x+2iy+3i=2z+3i
→f(z)=z²+3iz+C,由f(0)=-i得C=-i
∴f(z)=z²+3iz-i
∂v/∂x=2y+3
∂u/∂x=∂v/∂y=2x (柯西黎曼方程)
f‘(z)=∂u/∂x+i(∂v/∂x)=2x+2iy+3i=2z+3i
→f(z)=z²+3iz+C,由f(0)=-i得C=-i
∴f(z)=z²+3iz-i
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