I have the complex number (1 - E^2) (2 + I 2 Pi k) with k a natural number. How can I get the general formula for the argument? It should be Pi + ArcTan[k Pi].

I tried with:

Assumptions -> k > 0
Arg[(1 - E^2) (2 + 2 I Pi k)]

but it does not work.

Thank you for your help.


The following gets close:

    Arg[(1-E^2) (2+2 I Pi k)],
    TargetFunctions->{Re, Im}

ArcTan[-2, -2 k π]

  • $\begingroup$ Hi @CarlWoll. What does the comma mean? $\endgroup$ Sep 5 '17 at 21:56
  • 1
    $\begingroup$ @GennaroArguzzi Look it up here. Basically, ArcTan[x, y] is the same as ArcTan[y/x] except that the signs of x and y are used to determine the appropriate quadrant. $\endgroup$
    – Carl Woll
    Sep 5 '17 at 22:03
  • 1
    $\begingroup$ @Gennaro, you might meet it a lot of times in future work, so you might as well learn about two-argument arctangent. $\endgroup$
    – J. M.'s torpor
    Sep 6 '17 at 0:02
  • $\begingroup$ @CarlWoll the only strange thing is that real part $\frac{1-e^2}{2+2\pi^2 k^2}<0$ and imaginary part $\frac{\pi k(e^2-1)}{2+2\pi^2 k^2}>0$, thus I expect the result Pi+ArcTan[-2,-2 k Pi]. $\endgroup$ Sep 6 '17 at 7:13
  • 1
    $\begingroup$ @Gennaro, read the last sentence of my previous comment a bit more carefully and slowly. Then evaluate FullSimplify[ComplexExpand[Arg[a + b I] == ArcTan[a, b]], {a, b} ∈ Reals] and look at the result for a full minute. $\endgroup$
    – J. M.'s torpor
    Sep 6 '17 at 20:46

Try this:

Simplify[ComplexExpand[Arg[(1 - E^2)*(2 I Pi k)]], k > 0]

(*  -(\[Pi]/2)  *)

Have fun!


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