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Consider the equation

$$f'(x)+ g(x)f(x)=0$$

This equation is an ODE and has a solution $$ f(x)=C e^{ \int_1^x g(x) \ dx}.$$

Similarly, we can look at complex variables and consider the equation and Wirtinger derivatives

$$ (\partial_{\bar z} f)(z) +g(z) f(z)=0.$$

Can one still write down an explicit solution?

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  • $\begingroup$ Is $f(z)$ analytic? If so then the derivative of $f(z)$ with respect to $\bar{z}$ does not exist unless $f(z)$ is a constant. $\endgroup$ – Kevin Sep 23 at 13:59
  • $\begingroup$ No, it is not analytic... $\endgroup$ – Sascha Sep 23 at 14:06
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You can start by looking at the chain rule for wirtinger derivatives, from which you deduce that

$$ \partial_{\bar z} \exp(h(z)) = \exp(h(z)) \cdot \partial_{\bar z} h(z) $$

Therefore, if you find a function $h$ such that $\partial_{\bar z} h = - g(z)$ (I think you forgot a "$-$" sign in your solution for the real case!) taking $f(z) = \exp(h(z)) $ will solve your problem. In general, this is known as the d-bar problem (or $\bar\partial-$problem). As Daniele points out this Q&A is a god resource for the $\bar\partial-$problem in 1 dimension.

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    $\begingroup$ @Sascha, Jaume for a solution of the $1$-dim $\bar\partial$-problem this Q&A may be relevant. $\endgroup$ – Daniele Tampieri Sep 23 at 14:49
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    $\begingroup$ Indeed @daniele-tampieri! That's a way better link to the $\bar \partial$ problem than the wikipedia page that has (surprisingly?) almost no reference $\endgroup$ – Jaume Sep 23 at 15:01

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