If $1,A_1,A_2,A_3....A_{n-1}$ are the $n^{th}$ roots of unity then prove that $$\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}$$

What I did: I tried to use some of the following formulas: $$1+ A_1 +A_2+A_3+\cdots+A_{n-1} = 0$$

$$\dfrac{2^n - 1}{2-1} = (2 -A_1)(2-A_2)\cdots(2-A_{n-1})$$

and the fact that $|A_i| = 1$ for $i =1,2,3,\cdots,n-1$.


marked as duplicate by lab bhattacharjee, MR_BD, José Carlos Santos, ancientmathematician, Namaste Dec 17 '18 at 13:39

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  • $\begingroup$ Possible duplicate of Series involving complex roots. $\endgroup$ – Martin R Dec 17 '18 at 8:46
  • $\begingroup$ I am very sorry about it @ambretteorrisey $\endgroup$ – Smit Patel Dec 17 '18 at 9:43
  • $\begingroup$ @Smit Patel -- It's really not anywhere near being a matter of such magnitude ... and besides - you fixed it, I noticed, within a few minutes of my posting my comment. ¶ In fact, I'll delete the original one. ¶ Rather, I am very grateful to you for showing me so fascinating a theorem! ¶ And I do wonder whether it might be extensible to n>2. Not exactly the same theorem - but a more general one of which yours is a special case. $\endgroup$ – AmbretteOrrisey Dec 17 '18 at 9:47

Note that $$ z^n-1=\prod_{k=0}^{n-1}(z-a_k) $$ so that $$ nz^{n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k}\prod_{k=0}^{n-1}(z-a_k) $$ and therefore, $$ \frac{nz^{n-1}}{z^n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k} $$ Thus, $$ \sum_{k=0}^{n-1}\frac1{2-a_k}=\frac{n2^{n-1}}{2^n-1} $$ Subtracting the $k=0$ ($a_0=1$) term gives $$ \sum_{k=1}^{n-1}\frac1{2-a_k}=\frac{(n-2)2^{n-1}+1}{2^n-1} $$

  • $\begingroup$ @ robjohn can you please explain me how to bring second step for first step of your process $\endgroup$ – Smit Patel Dec 17 '18 at 9:41
  • $\begingroup$ @Smit Patel -- it's just the differentiation wrt $z$ of both sides of the equation ... if that is indeed what you are referring to. ¶ And this proof also well answers in the affirmative my speculation as to whether the theorem is extensible to numbers other than 2! The general case would be $${\big(n-{z\over z-1}\big)z^{n-1}+{1\over z-1}\over z^n-1}$$ ... but it loses its elegance for n≠2! $\endgroup$ – AmbretteOrrisey Dec 17 '18 at 10:30
  • $\begingroup$ Excellent method. Thank you for replying $\endgroup$ – Smit Patel Dec 17 '18 at 15:56
  • $\begingroup$ I use a variation of this function in other answers. Dividing $\frac{nz^{n-1}}{z^n-1}$ by $z^n$ leaves the residues at the roots of unity alone (each equals $1$), but the resulting function decays like $\frac1{|z|^{n+1}}$. That is, $\frac{n/z}{z^n-1}$ has a residue of $1$ at each $n^\text{th}$ root of unity, and a residue of $-n$ at $0$. See this answer, for example. $\endgroup$ – robjohn Dec 17 '18 at 18:31

Two ingredients for this proof:

  • The numbers $A_k$ for $1\leqslant k<n$ and $A_n=1$ are the roots of the polynomial $X^n-1$.
  • If $Y=\frac1{2-X}$ then $X=\frac{2Y-1}Y$.

Thus, the numbers $B_k=\frac1{2-A_k}$ for $1\leqslant k<n$ and $B_n=\frac1{2-A_n}=1$ are the solutions of the equation $$\left(\frac{2Y-1}Y\right)^n-1=0$$ that is, the roots of the polynomial $$(2Y-1)^n-Y^n$$

The highest degree terms of this polynomial are $$(2^n-1)Y^n-n2^{n-1}Y^{n-1}+\ldots$$ hence the sum of its roots is $$\sum_{k=1}^nB_k=1+\sum_{k=1}^{n-1}\frac1{2-A_k}=\frac{n2^{n-1}}{2^n-1}$$ from which the desired result follows.

  • $\begingroup$ Yepo! That's slick that is! Got two different proofs of this now. $\endgroup$ – AmbretteOrrisey Dec 17 '18 at 10:40

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