I don't know if there's any easy way to do it. Really it should be happening in the integration routine (that's why I said that this can be considered a bug). In the integration routine it has enough information to easily give an arctangent instead of a logarithm, but after the fact it's more difficult.
Aaron Meurer On Fri, Nov 11, 2016 at 8:44 AM, qm <[email protected]> wrote: > Am Freitag, 11. November 2016 10:55:49 UTC+1 schrieb qm: >> >> Do you know what kind of simplification I would have to apply to either of >> the two expressions r1,r2, to make SymPy actually also tell me that? > > > by 'simplification' I don't mean that I would have to substitute things like > log(a+1j) = log(1+a**2) + 1j*atan(1/a) by hand, but if there is a builtin > function to do that? > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To post to this group, send email to [email protected]. > Visit this group at https://groups.google.com/group/sympy. > To view this discussion on the web visit > https://groups.google.com/d/msgid/sympy/de5d5398-f440-4b39-bf0b-67201d873f49%40googlegroups.com. > > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CAKgW%3D6%2Bn8nGP8g_6LymuXfAWLonrvXVVz%2BAeUxRhhbPf8Ngz0g%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
