Why are $u(z)$ and $u(\bar{z})$ simultaneously harmonic? I'm trying to learn a bit of complex analysis, and this idea has got me stuck. 
I would like to show that, for $u$ a function of a complex variable $z$, that $u(z)$ and $u(\bar{z})$ are simultaneously harmonic.
I try writing $u(z)=a(z)+ib(z)$. Assuming $u(z)$ is harmonic, 
$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.
$$
Also, I think
$$
\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 a}{\partial x^2}+i\frac{\partial^2 b}{\partial x^2},
\qquad
\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 a}{\partial y^2}+i\frac{\partial^2 b}{\partial y^2}.
$$
I don't understand how to use this to show $u(z)$ and $u(\bar{z})$ are simultaneously harmonic. Aren't these the same function $u$? Shouldn't that be independent of whether you plug in $z$ or $\bar{z}$? Thanks.
 A: The Cauchy-Riemann equations are sufficient to show that $u$ is harmonic. Define $v(x,y)=u(\bar{z})$.
Now $z\ne \bar{z}$ so there's no reason to expect $u(z)=u(\bar{z})$ (take $u(z)=z$ and $z=i$ for instance), so we are not looking at the same function with $v(x,y)$. Note that $(\partial/\partial x)^2 v=v_{xx}=u_{xx}$ but using the chain rule gives:
$$v_{yy}=\frac{\partial}{\partial y}\left(-u_y\right(\bar{z}))=(-1)^{2}u_{xx}(\bar{z})$$
whence the harmonicity of $u(z)$ entails that of $u(\bar{z})$. (Keep in mind I abused the distinction between $\mathbb{C}$ and $\mathbb{R}^2$ quite severely here.)
A: Just use the chain rule. Note first that $u(\overline{z}) = u\circ\overline{\cdot} = u(x,-y)$.  The $x$ derivative of $u\circ\overline{\cdot}$ is the same as the $x$ derivative of $u$ and the $y$ derivative picks up a factor of $-1$ has $(u\circ\overline{\cdot})_y = -u_y$.  The second $y$ derivative picks up another $-1$, which gives $(u\circ\overline{\cdot})_{xx} + (-1)^2(u\circ\overline{\cdot})_{yy} = u_{xx} + u_{yy} = 0.$
