I've found more solutions. For `n=4` , note that `(1+i)^4-(1-i)^4=0`, and `(-1+i)^4-(-1-i)^4=0`, so `x=+-1` are solutions when `n=4.` Think of the geometry of the situation in the complex plane. When raising a complex number to a power `n`, you raise its magnitude to the `nth` power and multiply the angle it makes with the positive x-axis by `n`. If `x` is a real number, then `x+i` and `x-i` have the same magnitude (they're conjugates), so all we need to worry about are the angles. I found the above solutions easily by making complex numbers with angles of `45^o` and `-45^o`. When multiplied by 4 (I picked `n=4` for this reason), we end up with angles of `180^o` and `-180^0`, which correspond to the same complex number since we already know the magnitudes are equal.

We can do domething similar for any `n.` For `n=3`, we can pick a real number `x` such that `x+i` makes an angle of `180/3=60^o` with the positive x-axis. Since the complex part of our number is fixed at `i`, we already have our "opposite" side equal to 1, and this now becomes a basic trig problem. The "adjacent" side, which is `x`, must satisfy

`tan 60^o=1/x,` so `x=1/sqrt(3)`.

You can check that

`(1/sqrt(3)+i)^3-(1/sqrt(3)-i)^3=0.` These aren't the only ones (again, `+-x` are both solutions) and at this point I'm not sure if I have every solution for all `n`, but they're there and can be found by this method.

You should use binomial expansion such that:

`(x+i)^n = C_n^0 x^n + C_n^1 x^(n-1)*i + C_n^2 x^(n-2)*i^2 + C_n^3 x^(n-3)*i^3 + ... + C_n^n i^n`

`(x-i)^n = C_n^0 x^n- C_n^1 x^(n-1)*i + C_n^2 x^(n-2)*i^2- C_n^3 x^(n-3)*i^3 + ... + (-1)^n*C_n^n i^n`

`(x+i)^n - (x-i)^n = C_n^0 x^n + C_n^1 x^(n-1)*i + C_n^2 x^(n-2)*i^2 + C_n^3 x^(n-3)*i^3 + ... + C_n^n i^n - C_n^0 x^n+ C_n^1 x^(n-1)*i- C_n^2 x^(n-2)*i^2+ C_n^3 x^(n-3)*i^3 + ... + (-1)^n*C_n^n i^n`

`(x+i)^n - (x-i)^n = 2(C_n^1 x^(n-1)*i + C_n^3 x^(n-3)*i^3 + ....)`

But `(x+i)^n - (x-i)^n = 0 =>2(C_n^1 x^(n-1)*i + C_n^3 x^(n-3)*i^3 + ....) = 0`

You also should remember that `i^2 = -1, i^3 = -i , i^4 = 1` such that:

`2(C_n^1 x^(n-1)*i + C_n^3 x^(n-3)*i^3 + C_n^5 x^(n-5)*i^5 + C_n^7 x^(n-7)*i^7 + ....) = 0`

`C_n^1 x^(n-1)*i- C_n^3 x^(n-3)*i + C_n^5 x^(n-5)*i - C_n^7 x^(n-7)*i + ... = 0`

`i*(C_n^1 x^(n-1) + C_n^5 x^(n-5) + C_n^9 x^(n-9) + ...) - i*(C_n^3 x^(n-3) + C_n^7 x^(n-7) + ....) = 0`

`i*(C_n^1 x^(n-1) + C_n^5 x^(n-5) + C_n^9 x^(n-9) + ...)= i*(C_n^3 x^(n-3) + C_n^7 x^(n-7) + ....) `

`C_n^1 x^(n-1) + C_n^5 x^(n-5) + C_n^9 x^(n-9) + ... = C_n^3 x^(n-3) + C_n^7 x^(n-7) + ...`

**This equation holds only for x = 0, hence, evaluating the solution to the given equation yields x = 0.**

Ok, so I've spent more time with the problem. All solutions have to be real numbers. If `x` is complex but not real, then `x+i` and `x-i` have different magnitudes, so it's impossible for `|x+i|^n=|x-i|^n`, which must be the case if `x` is a solution.

Now for a given `n`, let `N=180^o/n`. Then

`1/tan(N)` , `1/tan(2N)` , `1/tan(3N)` ,..., `1/tan([n-1]N)` are all ` ` solutions.

There are `n-1` solutions here and the degree of the equation is also `n-1`, so these are the only solutions (which is another way to see that all solutions are real).

Again, these all work because of the way complex number exponentiation affects the angles of numbers written in polar form.