91Ó°ÊÓ

We need to check whether \(u(x, y)\) satisfies Laplace's equation, i.e., \(\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\). First, we find the partial derivatives of \(u(x, y)\). $$ \begin{aligned} \frac{\partial u}{\partial x} &= 2(1 - y) \\ \frac{\partial u}{\partial y} &= -2x \end{aligned} $$ Now, we find the second partial derivatives of \(u(x, y)\). $$ \begin{aligned} \frac{\partial^2 u}{\partial x^2} &= 0 \\ \frac{\partial^2 u}{\partial y^2} &= 0 \end{aligned} $$ As \(\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\), the function \(u(x, y)\) is harmonic.

We need to find a function \(v(x, y)\) such that it satisfies the Cauchy-Riemann equations: $$ \begin{aligned} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{aligned} $$ From step 1, we already have the first partial derivatives of \(u(x, y)\). $$ \begin{aligned} \frac{\partial u}{\partial x} &= 2(1 - y) \\ \frac{\partial u}{\partial y} &= -2x \end{aligned} $$ Now we apply the Cauchy-Riemann equations: $$ \begin{aligned} \frac{\partial v}{\partial y} &= 2(1 - y) \\ \frac{\partial v}{\partial x} &= 2x \end{aligned} $$ Integrating each equation to find the conjugate harmonic function \(v(x, y)\): $$ \begin{aligned} v(x, y) &= \int 2(1 - y) dy = 2(1 - y)y + g(x) \\ v(x, y) &= \int 2x dx = x^2 + h(y) \end{aligned} $$ Combining both equations, we get \(v(x, y) = 2(1 - y)y + x^2\), where we have set the integration constants to zero.

Now, we have the harmonic function \(u(x, y) = 2x(1 - y)\) and its conjugate harmonic function \(v(x, y) = 2(1 - y)y + x^2\). To express their combination as a function of \(z\), substitute \(z = x + \mathrm{i}y\): $$ f(z) = u(x, y) + \mathrm{iv}(x, y) = 2x(1 - y) + \mathrm{i}(2(1 - y)y + x^2) $$ Thus, the combined function \(f(z)\) is: $$ f(z) = 2x(1 - y) + \mathrm{i}(2(1 - y)y + x^2) $$