# 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.

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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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Thanks, anon. Forgive me, but why is $v_{yy}=\frac{\partial}{\partial y}(-u_x(\bar{z}))...$? Should it be $v_{yy}=\frac{\partial}{\partial y}(-u_y(\bar{z}))=(-1)^2u_{yy}(\bar{z})$, or am I mistaken? – Dedede Jan 22 '12 at 3:30
@Dedede: Typo, sorry. – anon Jan 22 '12 at 3:43
No worries, thanks. So is it always justifiable to write a function of a complex variable as a function of two real variables as was done in defining $v(x,y)$? – Dedede Jan 22 '12 at 3:47
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 $$f:\mathbb{R}^2\to\mathbb{R}^2:(x,y)\mapsto (\iota\circ f)(x+iy).$$ – anon Jan 22 '12 at 3:54

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 $(u\circ\overline{\cdot})_{xx} + (-1)^2(u\circ\overline{\cdot})_{yy} = u_{xx} + u_{yy} = 0.$

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thank you, Neal. – Dedede Jan 22 '12 at 3:51