I have a series like this one:

series = a[1][v](r-rp) + a[2][v](r-rp)^2 + O[r-rp]^3

where a[n][v] are complex coefficients of the series which are functions of v and (r-rp) is a real number. I want to Conjugate[] the series. However, Mathematica produces only (whatever I tried):

Conjugate[a[1][v](r-rp) + a[2][v](r-rp)^2 + O[r-rp]^3]

while I would like to obtain:

Conjugate[a[1][v]](r-rp) + Conjugate[a[2][v]](r-rp)^2 + O[r-rp]^3

I have tried e.g. Conjugate[ComplexExpand[series,{a}]] and variations with no good.

What should I do?


2 Answers 2


First, note the InputForm of the series expression:

series //InputForm

SeriesData[r, rp, {a[1][v], a[2][v]}, 1, 3, 1]

According to the documentation, and as you can see above, the 3rd argument consists of the coefficients of the series. So, to take the conjugate of a SeriesData object, we just need to map Conjugate at the right part:

conj = MapAt[Conjugate, series, 3];
conj //TeXForm

$(r-\operatorname{rp}) (a(1)(v))^*+(r-\operatorname{rp})^2 (a(2)(v))^*+O\left((r-\operatorname{rp})^3\right)$


Quick-and-dirty approach:

series = a[1][v] (r - rp) + a[2][v] (r - rp)^2 + O[r, rp]^3;
Assuming[r \[Element] Reals && rp \[Element] Reals,
  Conjugate /@ series // Simplify
  • $\begingroup$ Thank you for the answer. Could you explain to me, why for test = a+b Conjugate[test] returns Conjugate[a+b] while Conjugate /@ test returns Conjugate[a]+Conjugate[b]. $\endgroup$
    – fales
    Apr 5, 2018 at 16:29
  • 1
    $\begingroup$ @fales have a look at Map. $\endgroup$ Apr 5, 2018 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.