Holomorphic function $f=u+iv$ Let $f=u+iv$ with $u,v\in \mathcal C^2$ and $u$, $v$, $xu(x,y)-yv(x,y)$ and $xv(x,y)-yu(x,y)$ harmonic (i.e. the laplacian is nul). I have to prove that $f$ is holomorphic, but with no success.
 A: The answer was essentially given in the comments. Here are some details.
Let $u(x,y)=y$ and $v(x,y)=x$. Then, $u$, $v$, $xu(x,y)-yv(x,y)$ and $xv(x,y)-yu(x,y)$ are all harmonic. However, $f=u+\mathbf{i}v$ is not holomorphic because
$$u_y=1\neq -1=-v_x.$$
So, with your assumptions, is not possible to prove that $f$ is holomorphic.
Notice that if $u$, $v$, $xu(x,y)-yv(x,y)$ and $xv(x,y)+yu(x,y)$ are harmonic, then $f=u+\mathbf{i}v$ is holomorphic. In fact, in this case,
\begin{align*}
0&=\Big(xu(x,y)-yv(x,y)\Big)_{xx}+\Big(xu(x,y)-yv(x,y)\Big)_{yy}\\
&=2u_x(x,y)-2v_y(x,y)
\end{align*}
and
\begin{align*}
0&=\Big(xv(x,y)+yu(x,y)\Big)_{xx}+\Big(xv(x,y)+yu(x,y)\Big)_{yy}\\
&=2v_x(x,y)+2u_y(x,y).
\end{align*}
Thus, $$u_x=v_y,\quad u_y=-v_x.\tag{Cauchy-Riemann equations}$$
A: I'm assuming instead that $xv + yu$ is harmonic, otherwise the result is not true (consider $u = y$ and $v = 0$). Assuming this, we have 
$$zf = (xu - yv) + i(xv + yu)$$
is harmonic. Thus, letting $\partial_z = (1/2)(\partial_x - i\partial_y)$ and $\partial_{\bar{z}} = (1/2)(\partial_x + i\partial_y)$, by the product rule
$$0 = \partial_z\partial_{\bar{z}}(zf) = \partial_z(\underbrace{{\partial_{\bar{z}} z}}_{=\, 0} f + z\partial_{\bar{z}} f)) = \partial_{\bar{z}}f + z\partial_z\partial_{\bar{z}}f.\tag{*}$$
Since $u$ and $v$ are harmonic, $f$ is harmonic, so $\partial_z\partial_{\bar{z}}f = 0$. Hence by $(*)$, we have $\partial_{\bar{z}} f = 0$. Therefore, $f$ holomorphic.
