Real and imaginary part of an analytic function are harmonic? Let $f:\Bbb{C}\to\Bbb{C}$ be an analytic function. For $z = x+iy,$ let $u,v : \mathbb{R}^2  \rightarrow \mathbb{R}^2 $ be such that
$u(x,y) = \operatorname{Re}f(z)$ and $v(x,y) = \operatorname{Im}f(z)$.
Then which of the following are correct,
1.$\dfrac{\partial ^2u}{\partial x^2} + \dfrac{\partial ^2u}{\partial y^2} = 0$
2.$\dfrac{\partial ^2v}{\partial x^2} + \dfrac{\partial ^2v}{\partial y^2} = 0$
3.$\dfrac{\partial ^2u}{\partial x \partial y} - \dfrac{\partial ^2u}{\partial y \partial x} = 0$
My try: Since $\operatorname{Re} f(z)$ and $\operatorname{Im}f(z)$ parts of an analytic function satisfies Cauchy Riemann Equation. We can conclude all the options holds.
But it's displayed option $2$ is incorrect. Is that so?

If $f = u + iv$ then $f' = u_x + i v_x = {1 \over i} (u_y+i v_y)$ and hence
$f'' = u_{xx} + i v_{xx} = -(u_{yy} + i v_{yy})$. Then 1 & 2 follow from this.
Similarly, $f'' = {1 \over i} (u_{xy}+ i v_{xy}) = {1 \over i} (u_{yx}+ i v_{yx}) $ and 3 follows from this.
