Question: Suppose $u,v$ are harmonic and satisfy the Cauchy-Riemann equations in $\mathbb{R}^2$. Show that $f = u + iv$ satisfies $f'(x) = u_x(x,0) - iu_y(x,0)$ for real $x$.
My attempts: I considered the fact that $\displaystyle \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ and from there came to the conclusion that $\displaystyle \frac{\partial ^2 u}{\partial x^2} = -\frac{\partial ^2 u}{\partial y^2}$ and so $\displaystyle \frac{\partial u}{\partial x} = -\frac{\partial u}{\partial y} + C$. I'm not entirely sure where to go from here. Is this on the right track? Any help or guidance would be greatly appreciated!
We have $u,v \in C^2( \mathbb R^2,\mathbb R)$, hence $u$ and $v$ are real differentiable, thus $f$ is real differentiable. Since $u,v$ satisfy the Cauchy-Riemann equations , $f$ is holomorphic on $ \mathbb C.$ Hence
$$f'(z)=u_x(x,y)+iv_x(x,y)$$
for all $z=x+iy \in \mathbb C.$ From $v_x=-u_y$ we get
$$f'(x) = u_x(x,0) - iu_y(x,0)$$
for $x \in \mathbb R.$