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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.

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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.)

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    Yes. To be technical, let $f:\mathbb{C}\to\mathbb{C}$. Then $\iota:\mathbb{C}\to\mathbb{R}^2:z\mapsto(\mathrm{Re}\; z,\mathrm{Im}\; z)$ allows us to define $f:\mathbb{R}^2\to\mathbb{R}^2:(x,y)\mapsto (\iota\circ f)(x+iy).$2019-05-04
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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 $(u\circ\overline{\cdot})_{xx} + (-1)^2(u\circ\overline{\cdot})_{yy} = u_{xx} + u_{yy} = 0.$