九、设 f(z)=u+iv 解析,且 -v=(x-y)((x)^2+4xy+(y)^2), 求f(z)·

步骤 1：求偏导数
对 $u-v=(x-y)({x}^{2}+4xy+{y}^{2})$ 分别对 $x$ 和 $y$ 求偏导数，得到：
$$
\frac{\partial (u-v)}{\partial x} = 3x^2 + 6xy - 3y^2
$$
$$
\frac{\partial (u-v)}{\partial y} = 3x^2 - 6xy - 3y^2
$$
步骤 2：应用柯西-黎曼方程
根据柯西-黎曼方程，有：
$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}
$$
$$
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
$$
步骤 3：解偏微分方程
联立上述方程，得到：
$$
\frac{\partial v}{\partial y} - \frac{\partial v}{\partial x} = 3x^2 + 6xy - 3y^2
$$
$$
-\frac{\partial v}{\partial x} - \frac{\partial v}{\partial y} = 3x^2 - 6xy - 3y^2
$$
解得：
$$
\frac{\partial v}{\partial x} = 3y^2 - 3x^2
$$
$$
\frac{\partial v}{\partial y} = 6xy
$$
步骤 4：求解 $v$
积分得到：
$$
v = 3x^2y - x^3 + \varphi(y)
$$
$$
\frac{\partial v}{\partial y} = 6xy + \varphi'(y) = 6xy
$$
$$
\varphi'(y) = 0
$$
$$
\varphi(y) = C_1
$$
$$
v = 3x^2y - x^3 + C_1
$$
步骤 5：求解 $u$
同理可得：
$$
\frac{\partial u}{\partial x} = 6xy
$$
$$
\frac{\partial u}{\partial y} = 3x^2 - 3y^2
$$
积分得到：
$$
u = 3x^2y - y^3 + \varphi(x)
$$
$$
\frac{\partial u}{\partial x} = 6xy + \varphi'(x) = 6xy
$$
$$
\varphi'(x) = 0
$$
$$
\varphi(x) = C_2
$$
$$
u = 3x^2y - y^3 + C_2
$$
步骤 6：求解 $f(z)$
$$
f(z) = u + iv = 3x^2y - y^3 + C_2 + i(3x^2y - x^3 + C_1)
$$
令 $z = x + yi$，则有：
$$
f(z) = z^3 + C(1 + i)
$$