# Get argument formula of a complex number with integer variable?

GROUPS:
 Hello everyone. I have the complex number (1 - E^2) (2 + I 2 Pi k) with k integer. How can I get the general formula for the argument? It should be Pi+ArcTan[k Pi] if k>0.I tried with: Assumptions->Element[k,Integers] Assumptions -> k > 0 Arg[(1 - E^2) (2 I Pi k)] but it does not work.Thank you very much.
 This gives an idea In[18]:= Arg[-2 - I # \[Pi]] & /@ Range[10] Out[18]= {-\[Pi] + ArcTan[\[Pi]/2], -\[Pi] + ArcTan[\[Pi]], -\[Pi] + ArcTan[(3 \[Pi])/2], -\[Pi] + ArcTan[2 \[Pi]], -\[Pi] + ArcTan[(5 \[Pi])/2],-\[Pi] + ArcTan[3 \[Pi]], -\[Pi] + ArcTan[(7 \[Pi])/2], -\[Pi] + ArcTan[4 \[Pi]],-\[Pi] + ArcTan[(9 \[Pi])/2], -\[Pi] + ArcTan[5 \[Pi]]} but here you see a rational positive k does it as well In[24]:= FindInstance[Arg[-2 - I k \[Pi]] == -\[Pi] + ArcTan[k \[Pi]/2], {k}] Out[24]= {{k -> 106/5}} why not compute it directly, it's just plane trigonometry.
 Because of Numeric Quantities and Arg Arg[z] is left unevaluated if z is not a numeric quantity. one has In[15]:= Arg[-2 - I GoldenRatio \[Pi]] Out[15]= -\[Pi] + ArcTan[(GoldenRatio \[Pi])/2] but (because x is not a numeric quantity) In[31]:= Simplify[Arg[-2 - I \[Pi] x], x \[Element] Reals && x > 0] Out[31]= Arg[-2 - I \[Pi] x] the assumption x > 0 does not constitute x as numeric quantity.
 Daniel Lichtblau 2 Votes Could use Refine in tandem with ComplexExpand. The latter assumes variables take on real values unless instructed otherwise. Refine[ ComplexExpand[Arg[(1 - E^2) (2 + I 2 Pi k)], TargetFunctions -> {Re, Im}], k \[Element] Reals && k > 0] (* Out[624]= -\[Pi] + ArcTan[k \[Pi]] *)