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Regression of complex-valued functions

On 13/02/14 12:03, Andrea Graziani wrote:
<SNIP>
I find it a bit surprising that there are differences in the third 
decimal place here.  I would have thought that the results would be 
exactly the same, or ***very*** close to it.

If you have done it right (I have not checked your code at all) the "two 
approaches" should amount to the same approach.  I.e. Mod(z)^2 is
equal to Re(z)^2 + Im(z)^2.

Check your code carefully.

Also it would be a good idea to try a "toy example", something much less 
complicated than your real problem, and see what happens there.

cheers,

Rolf Turner