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Get argument formula of a complex number with integer variable?

Posted 1 year ago
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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 -> k > 0
Arg[(1 - E^2) (2 I Pi k)]

but it does not work.

Thank you very much.

3 Replies

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.

Could use Refine in tandem with ComplexExpand. The latter assumes variables take on real values unless instructed otherwise.

ComplexExpand[Arg[(1 - E^2) (2 + I 2 Pi k)], 
TargetFunctions -> {Re, Im}], k \[Element] Reals && k > 0]

(* Out[624]= -\[Pi] + ArcTan[k \[Pi]] *)
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