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I have an expression of the form $(\frac{-1+e^{i\,N\,\phi}}{-1+e^{i \phi}})^n$.
I want to use mma to hopefully simplify it.

I have tried ComplexExpand[], PowerExpand[] with simplifying assumption in Refine[] that $N,\phi$ are real. But on using Simplify[] or FullSimplify[]. Either mma does nothing or gives a complicated answer.

a = PowerExpand[\!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(N - 1\)] \*SuperscriptBox[\(E\), \((I\ k\ \[Phi])\)]\)] FullSimplify[ComplexExpand[a^n]] Refine[FullSimplify[PowerExpand[a^n]], Assumptions -> {N \[Element] Reals, \[Phi] \[Element] Reals}] ```
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  • $\begingroup$ Welcome to Mathematica StackExchange! You have to be careful. Some symbols and names have an internal meaning. N[] is an internal symbol so cannot be used. $\endgroup$ Commented Apr 25, 2022 at 11:07

1 Answer 1

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Maybe:

 Func = Simplify[((-1 + E^(I M \[Phi]))/(-1 + E^(I \[Phi])) // ExpToTrig // TrigFactor // FullSimplify)^n, Assumptions -> {M \[Element] Reals, \[Phi] \[Element] Reals, n > 0}] // PowerExpand ComplexExpand[Re[Func[[1]]], TargetFunctions -> {Re, Im}]*Func[[2 ;; 3]](*Real part*) (*Cos[1/2 (-1 + M) n \[Phi]] Csc[\[Phi]/2]^n Sin[(M \[Phi])/2]^n*)

$$\left(\sum _{k=0}^{M-1} e^{i k \phi }\right){}^n=\cos \left(\frac{1}{2} (-1+M) n \phi \right) \csc ^n\left(\frac{\phi }{2}\right) \sin ^n\left(\frac{M \phi }{2}\right)$$

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  • $\begingroup$ Thank you so much for timely answer. I was hoping that I would get a form in which the power n comes inside the argument of cos and sines. That would have helped simplify my original calculations. $\endgroup$ Commented Apr 26, 2022 at 11:11
  • $\begingroup$ @HafizmuhammadIbrahimjaffar Sin[x]^n can't be better simplified. $\endgroup$ Commented Apr 26, 2022 at 14:53

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