# Simplifying a power of complex exponential quotient

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}]
$$$$

• 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. Apr 25, 2022 at 11:07

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*)


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

• 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. Apr 26, 2022 at 11:11
• @HafizmuhammadIbrahimjaffar Sin[x]^n` can't be better simplified. Apr 26, 2022 at 14:53