Consider the following simple problem.
Let $f$ be a complex analytic function over $\mathbb{R}$ such that $f(x) in \mathbb{R}$ for each $x \in \mathbb{R}$. Prove that $f(z) = \overline{f(\overline{z})}$.
Of course this can be solved by observing that all the coefficients of the Maclaurin series are reals and the convergence valid over $\mathbb{C}$.
One can also approach the proof by defining $f(z) - \overline{f(\overline{z})}$ and observing the zeroset contains $\mathbb{R}$.
Suppose that we are trying to solve this with real analysis techniques. So let $f(x+yi) = u(x,y)+iv(x,y)$ where $x,y \in \mathbb{R}$ and $u$, $v$ are real valued. Then the problem reduces to the following question in differential equations. I'm looking for some calculus level argument which solves the following.
Let $u(x,y)$ and $v(x,y)$ be real valued functions which satisfies the Cauch-Riemann equation $u_x = v_y$, $u_y = -v_x$ on $\mathbb{R}^2$. Suppose that $v(x,0) =0$ for all $x \in \mathbb{R}$. Prove that $u(x,y) = u(x,-y)$ and $v(x,y) = -v(x,-y)$.
If we let $U(x,y) = u(x,y) - u(x,-y)$ and $V(x,y) = v(x,y) +v(x,-y)$. One can also approach this by proving
If $U,V$ are real valued functions which satisfies the Cauchy-Riemann equation. Suppose that $U(x,0) = V(x,0) = 0$. Then $U$, $V$ vanishes on $\mathbb{R}^2$.
Thank you for your attention.