Find all holomorphic functions $f(x + iy) = u(x, y) + iv(x, y)$, for which it holds everywhere that
$$ u(x, y) + v(x, y) = x^2 – y^2. $$
Express the solution as a function of the variable $z = x + iy$.
Attempt: To find the holomorphic functions $f(z) = u(x, y) + iv(x, y)$ where $z = x + iy$ and $u(x, y) + v(x, y) = x^2 – y^2$, we start by recognizing that for $f$ to be holomorphic, $u$ and $v$ must satisfy the Cauchy-Riemann equations: $u_x = v_y$ and $u_y = -v_x$. Given the condition $u + v = x^2 – y^2$, we differentiate this equation with respect to $x$ and $y$. Differentiating with respect to $x$, we get $u_x + v_x = 2x$, and differentiating with respect to $y$, we get $u_y + v_y = -2y$. Using the Cauchy-Riemann equations, $u_x = v_y$ and $u_y = -v_x$, we can substitute to find $v_y + v_x = 2x$ and $-v_x + v_y = -2y$. Solving these simultaneous equations, we get $v_y = x – y$ and $v_x = x + y$. Integrating these partial derivatives, we find $v = xy + C(y)$ and $v = \frac{y^2}{2} + C(x)$. Equating the two expressions for $v$ and solving for the arbitrary functions, we determine $C(y) = -\frac{y^2}{2}$ and $C(x) = \frac{x^2}{2}$. Therefore, the function $f(z) = z^2 – i \left( \frac{z^2}{2} \right) = z^2 – \frac{i z^2}{2}$, but this clearly doesn't satisfy that equation, so where did I make an error?
Best Answer
The problem can be solved without applying the C-R equations.We will make use of the fact that the real part of holomorphic function determines the imaginary part up to a real additive constant.
The real part of the holomorphic function $(1-i)f(z)$ is equal $u+v=x^2-y^2.$ Following the simplification suggested by @Martin R , the holomorphic function with real part $x^2-y^2 $ is of the form $z^2+ic, $ $c\in\mathbb{R}.$ Hence we get $$(1-i)f(z)=z^2+ic$$ for a real constant $c.$ Multiplying by ${1\over 2}(1+i)$ gives $$f(z) ={1\over 2}(1+i)z^2+{1\over 2}(i-1)c$$