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Can anyone help explain to me what the general solution is to this? As I understand it is not the same as dealing with PDE's as $f$ is a function of $z=x+iy$ so is probably not $f=g(z)+h(\bar{z})$

$$\frac{\partial}{\partial \bar{z}}\left(\frac{\partial f}{\partial z}\right)=0$$

Thanks very much in advance

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You may identify that $$ \Delta f = 4 \frac{\partial^{2} f}{\partial z \partial \bar{z}},$$ so that the equation in question reduces to the Laplace equation. Then the solutions are (by definition) harmonic functions. –  sos440 Dec 28 '11 at 14:03
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An extra note to @sos440's comment: if you write $x=(z+\bar{z})/2$ and $y=(z-\bar{z})/(2i)$ and use the chain rule, the relationship between your operator and the Laplacian should become clear. –  Chris Taylor Dec 28 '11 at 14:09
    
@sos440: Thanks very much, and so to Chris Taylor, that was me being slow –  Freeman Dec 28 '11 at 14:12
    
@sos440: So I see that is reduces to laplaces equation, but how do you go about solving that with a complex function? –  Freeman Dec 28 '11 at 14:17
    
could you say that this means that $u_{xx}+u_{yy}+i(v_{xx}+v_{yy})=0$, then solve for $u,v$ normally? –  Freeman Dec 28 '11 at 14:20
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1 Answer

up vote 2 down vote accepted

Your differential equation is solved by any holomorphic function.

In fact, call $w=\frac{\partial f}{\partial z}$ and your equation rewrites $\frac{\partial w}{\partial \bar{z}}=0$; hence, if the domain of $f$ (and of $w$, of course) is a simply connected open set, $w$ is a holomorphic function; but then also $f$ is holomorphic (for it is an antiderivative of $w$).

On the other hand, is $f$ is holomorphic in a simply connected open set, then $\frac{\partial f}{\partial z} =f^\prime$ is also holomorphic there; therefore $\frac{\partial f^\prime}{\partial \bar{z}} =0$.

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yes, but that's not the general solution. –  user20266 Dec 28 '11 at 17:04
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