This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory real analysis.

I request to help find the solution.

If $n$ is a positive integer, find all roots of the equation : $$(1+\frac{ix}n)^n = (1-\frac{ix}n)^n$$

The binomial expansion on each side will lead to:

$$(n.1^n+C(n, 1).1^{n-1}.\frac{ix}n + C(n, 2).1^{n-2}.(\frac{ix}n)^2 + C(n, 3).1^{n-3}.(\frac{ix}n)^3+\cdots ) = (n.1^n+C(n, 1).1^{n-1}.\frac{-ix}n + C(n, 2).1^{n-2}.(\frac{-ix}n)^2 + C(n, 3).1^{n-3}.(\frac{-ix}n)^3+\cdots )$$

$n$ can be odd or even, but the terms on l.h.s. & r.h.s. cancel for even $n$ as power of $\frac{ix}n$. Anyway, the first terms cancel each other.

$$(C(n, 1).1^{n-1}.\frac{ix}n + C(n, 3).1^{n-3}.(\frac{ix}n)^3+\cdots ) = (C(n, 1).1^{n-1}.\frac{-ix}n + C(n, 3).1^{n-3}.(\frac{-ix}n)^3+\cdots )$$

As the term $(1)^{n-i}$ for $i \in \{1,2,\cdots\}$ don't matter in products terms, so ignore them:

$$(C(n, 1).\frac{ix}n + C(n, 3).(\frac{ix}n)^3+\cdots ) = (C(n, 1).\frac{-ix}n + C(n, 3).(\frac{-ix}n)^3+\cdots )$$

$$2(C(n, 1).\frac{ix}n + C(n, 3).(\frac{ix}n)^3+\cdots ) = 0$$

$$C(n, 1).\frac{ix}n + C(n, 3).(\frac{ix}n)^3+\cdots = 0$$

Unable to pursue further.


2 Answers 2


Hint: Put $$z=\frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}$$ then $z$ will be a $n$-root of unity and solve for $x:$ $$z= \frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}=\exp{\left(i\frac{2k\pi}{n}\right)},\quad k\in\{0,1,...,n-1\}$$

  • 2
    $\begingroup$ There should be a Pi as well to be exact. $\endgroup$
    – plus1
    May 26, 2019 at 2:08
  • $\begingroup$ @plus1 A typo. Corrected $\endgroup$
    May 26, 2019 at 2:08

Let $1=r\cos t,\dfrac xn=r\sin t$

Using http://mathworld.wolfram.com/deMoivresIdentity.html

$ r^n(\cos nt+i\sin nt)=r^n(\cos nt-i\sin nt)$

$\iff \sin nt=0$

$\implies nt=k\pi$

$ \dfrac xn=\tan t=\tan\dfrac{k\pi}n$ where $0\le k\le n-1$

  • $\begingroup$ Please elaborate (geometrically, or otherwise) your answer's first line. It forms the core of the transformation of the question, but is not clear how arrived at it. If it is just an assumption, then what conditions it imposes on the solution? $\endgroup$
    – jiten
    May 26, 2019 at 3:03
  • 1
    $\begingroup$ @jiten, To utilize math.stackexchange.com/questions/1432568/…, $$a+ib$$ can be written as $$r(\cos t+i\sin t)$$ without any assumption $\endgroup$ May 26, 2019 at 3:13

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