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I am a PhD student and I have to address the optimization of a real scalar function of complex variables $f(z_1,z_2,\ldots,z_n)$. The function is real valued in my case because each complex $z_i$ comes with its complex conjugate, i.e. there is $z_j$, $j \neq i$ such that $z_i = \bar{z_j}$.

I know that an usual way to address such a problem is to perform the optimization with respect to the real and complex parts of the variables, but working in complex allows me to keep a convenient structure for the function and its gradient.

Yet the issue is that in my algorithm, numerical errors appear and tend to perturbate the fact that the optimization variables must come in conjugate pairs. Hence I would like to know if there is a convenient way to express this constraint without knowing a priori which points must be complex conjugate ?

For the time being, I keep an index list of the complex conjugate pairs and each time I compute the function, I check if they are still complex conjugate, if not, i adjust them the best I can. Yet it is not a generic approach and it is not very satisfactory. In particular, it becomes really tedious when the variables cross the real line.

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  • $\begingroup$ Obviously, the fact that for each $i$ there exists a $j$ with $z_i=\overline{z}_j$ doesn't imply that the function is real valued. So I'm not at all sure what you mean to say. $\endgroup$
    – Steven Landsburg
    Commented Oct 23, 2013 at 12:50
  • $\begingroup$ In fact, I meant that the function $f$ is a real valued function (by definition) of some complex variables $z$ and their complex conjugate $\bar{z}$, i.e. $f(z) = f(z,\bar{z})$. But in practice, $z$ and $\bar{z}$ are not separated entities and whenever the complex variables does not come with their complex conjugate, $f$ does not have the right meaning anymore, so I would like to force my variables to remain closed under conjugation. $\endgroup$
    – Pierre Vuillemin
    Commented Oct 23, 2013 at 13:12
  • $\begingroup$ I'm afraid this makes even less sense now than it did before your clarification. $\endgroup$
    – Steven Landsburg
    Commented Oct 23, 2013 at 13:26
  • $\begingroup$ Maybe it would be more clear with the function itself. The function $f$, is given by $ f(z_1,\ldots,z_n,c_1,\ldots,c_n) = \sum_{i=1}^n \sum_{j=1}^n \frac{c_i c_j}{z_i+z_j}$. The $c_i$, $i=1,\ldots,n$ are also complex variables which come with their conjugate, if $z_i = \bar{z_j}$ then $c_i = \bar{c_j}$, $i\neq j$. $\endgroup$
    – Pierre Vuillemin
    Commented Oct 23, 2013 at 13:39
  • $\begingroup$ So suppose you evaluate $f(z,\overline{z})$ at the point where $z=3+i$ and $\overline{z}=3-i$. Then you go back and check to see whether $3+i$ and $3-i$ are still conjugate. But I can tell you in advance that $3+i$ and $3-i$ will be conjugate till the end of time. So what are you actually checking? $\endgroup$
    – Steven Landsburg
    Commented Oct 23, 2013 at 13:49

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