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Knowing the expression of $\dot{z}$: \begin{equation} \dot{z}=\lambda z+ \frac{g_{20}}{2} z^2+g_{11} z \overline{z}+\frac{g_{02}}{2}\overline{z}^2, \ \ \ z,\lambda\in C \end{equation} How do I calculate $\dot{\overline{z}}$? Maybe it's a silly question, but I'm just starting to study complex analysis, self-taught.

Thank you very much

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what is $g$? A function, a complex number? –  Nameless Dec 8 '12 at 9:28
    
Sorry, yes is a complex number. –  Mark Dec 8 '12 at 9:31

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up vote 1 down vote accepted

I have the feeling that Nameless's answer is not correct (there is a-priory no reason why $z$ and $\bar z$ should fulfill the same differential equation). The OP wanted the solution for $\lambda\in\mathbb{C}$ (and I guess also $g$'s are complex numbers).

I will assume (please correct me if I am wrong) that $\dot z$ denotes $\partial_t z(t)$ and that $t$ is in fact a real number. In this case you can convince yourself that $$\bar {\dot z}=\overline{\partial_t z(t)} = \overline{\partial_t x(t)+ i \partial_t y(t)}= \partial_t x(t)- i \partial_t y(t)= \partial_t \bar z(t) =\dot {\bar z} ,$$ i.e., complex conjugation and the temporal derivative commute!

Therefore, you get the equation for $\dot {\bar z} $ just by taking the complex conjugate of your equation $$ \dot{\bar z}=\bar{\lambda} \bar{z}+ \frac{\bar{g}_{20}}{2} \bar{z}^2+\bar{g}_{11} \bar{z} z+\frac{\bar{g}_{02}}{2}z^2.$$

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