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解析要求满足柯西黎曼条件:∂u/∂x=∂v/∂y,∂u/∂y=-∂v/∂x
u=(x+y)/(x²+y²),v=(x-y)/(x²+y²)
∂u/∂x=[(x²+y²)-2x(x+y)]/(x²+y²)²=[(x²+y²)-2x(x+y)]/(x²+y²)²=(y²-x²-2xy)/(x²+y²)²
∂u/∂y=(x²-y²-2xy)/(x²+y²)²
∂v/∂x=(y²-x²+2xy)/(x²+y²)²
∂v/∂y=-(x²-y²+2xy)/(x²+y²)²
验证得:∂u/∂x=∂v/∂y,∂u/∂y=-∂v/∂x两个均恒成立,因此该函数在整个复平面解析,其导数为:
∂u/∂x+i∂v/∂x
=(y²-x²-2xy)/(x²+y²)² + i(y²-x²+2xy)/(x²+y²)²
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。
u=(x+y)/(x²+y²),v=(x-y)/(x²+y²)
∂u/∂x=[(x²+y²)-2x(x+y)]/(x²+y²)²=[(x²+y²)-2x(x+y)]/(x²+y²)²=(y²-x²-2xy)/(x²+y²)²
∂u/∂y=(x²-y²-2xy)/(x²+y²)²
∂v/∂x=(y²-x²+2xy)/(x²+y²)²
∂v/∂y=-(x²-y²+2xy)/(x²+y²)²
验证得:∂u/∂x=∂v/∂y,∂u/∂y=-∂v/∂x两个均恒成立,因此该函数在整个复平面解析,其导数为:
∂u/∂x+i∂v/∂x
=(y²-x²-2xy)/(x²+y²)² + i(y²-x²+2xy)/(x²+y²)²
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。
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