设f(z)=2x(1-y)+i(x²-y²+2y)，问函数f(z)在何处可导，并在可导处求出f'(z).

将函数 $ f(z) = 2x(1-y) + i(x^2 - y^2 + 2y) $ 分解为实部 $ u(x,y) = 2x(1-y) $ 和虚部 $ v(x,y) = x^2 - y^2 + 2y $。计算偏导数： \[ \frac{\partial u}{\partial x} = 2 - 2y, \quad \frac{\partial u}{\partial y} = -2x, \quad \frac{\partial v}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 2 - 2y \] 满足柯西-黎曼方程： \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] 故函数在复平面上处处可导。导数为： \[ f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = (2 - 2y) + i(2x) \] 或表示为： \[ \boxed{(2 - 2y) + i(2x)} \quad \text{或} \quad \boxed{2 + 2i\overline{z}} \]