Why can the complex conjugate of a variable be treated as a constant when differentiating with respect to that variable?

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    $\begingroup$ maybe this question can help math.stackexchange.com/questions/852978/meaning-of-fz-barz $\endgroup$ – mercio Nov 28 '15 at 19:30
  • $\begingroup$ What I'm really asking is if z = x+jy and zbar = x-jy. When z is differentiated with respect to any variable and nonzero, either x must change or y must change. But if either x or y changes, then zbar must also change, making it not possible to differentiate z with respect to given variable while keeping zbar constant with respect to that same variable. $\endgroup$ – chris Dec 2 '15 at 1:30

The Wirtinger derivative $\frac{\partial }{\partial z}$ is a derivation, meaning that it is linear in $f$ and satisfies the Leibniz rule $$ \frac{\partial }{\partial z}(fg)=f\frac{\partial }{\partial z}g + g \frac{\partial }{\partial z}f $$ (Easy to derive from the standard Leibniz rule.) In addition, a direct calculation shows that $$ \frac{\partial }{\partial z}\bar z=0 $$ Combined, the above facts imply that
$$ \frac{\partial }{\partial z}(\bar z f)=\bar z\frac{\partial }{\partial z}( f) $$ for any real-differentiable function $f$. Repeated implication of the above shows that $$ \frac{\partial }{\partial z}(\bar z^n f)=\bar z^n\frac{\partial }{\partial z}( f) $$ for every integer $n$. Finally, real-analytic functions can be differentiated by taking the derivative of their power series term-by-term, which justifies computations like $$ \frac{\partial }{\partial z}(e^{\bar z} f)=e^{\bar z}\frac{\partial }{\partial z}( f) $$


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