TITLE: Find all roots of the equation :$(1+\frac{ix}n)^n = (1-\frac{ix}n)^n$
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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 $0\le k\le n-1$