Maybe it is a very stupid question but I am having trouble with summing complex numbers in Mathematica. I have

q = l1 E^(2 π I t1) + l2  E^(2 π I t2)

where l1, l2 t1, and t2 are known real numbers. I obtain an answer which is a set of four complex numbers.

I think it sums all the components one by one (Minkowski sum?), and I want just to sum two complex numbers to obtain a third one. What function should I use?

  • 3
    $\begingroup$ You might want to look at ComplexExpand[]. $\endgroup$ Jul 11, 2015 at 12:29
  • $\begingroup$ What do you mean by "2[Pi]"? $\endgroup$
    – Peltio
    Jul 11, 2015 at 12:30
  • $\begingroup$ @Peltio, Markdown strips backslashes, unfortunately. I've fixed it. $\endgroup$ Jul 11, 2015 at 12:37
  • $\begingroup$ q // ExpToTrig // Simplify is equivalent to q//ComplexExpand in this case. $\endgroup$
    – Bob Hanlon
    Jul 11, 2015 at 13:16

2 Answers 2


Since l1, l2, t1, t2 are known, you just need to plug them in:

q=l1 E^(2 \[Pi] I t1) + l2 E^(2 \[Pi] I t2) /. {l1->0.2, t1->0.1, l2->5, t2->-1}

gives 5.1618 + 0.117557 I

   q1 = l1 E^(2 \[Pi] I t1) + l2 E^(2 \[Pi] I t2)

        q2 = q1 // ExpToTrig

        (Drop[q2, 2] // Factor) + Drop[q2, -2]

        (*  E^(2 I \[Pi] t1) l1 + E^(2 I \[Pi] t2) l2

        l1 Cos[2 \[Pi] t1] + l2 Cos[2 \[Pi] t2] + I l1 Sin[2 \[Pi] t1] + 
         I l2 Sin[2 \[Pi] t2]

        l1 Cos[2 \[Pi] t1] + l2 Cos[2 \[Pi] t2] + 
         I (l1 Sin[2 \[Pi] t1] + l2 Sin[2 \[Pi] t2])  *)

Have fun!


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.