real analysis - Find all roots of the equation :$(1+frac{ix}n)^n = (1-frac{ix}n)^n$

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.

Answer

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 $x:$ $$z= \frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}=\exp{\left(i\frac{2k\pi}{n}\right)},\quad k\in\{0,1,...,n-1\}$$