If (x + 1)n = xn, then |x + 1|n = |x|n, and hence |x + 1| = |x|. Geometrically, |x + 1| = |x| means that the point x in the complex plane is the line segment bisector of the segment |AB| with A(-1, 0) and B(0, 0), and this is actually the line Re(x) = -1/2 (d(x, A) = d(x, B)). (Also, the equality |x + 1|2 = |x|2 immediately yields Re(x) = -1/2).

Dear Romeo,

Your solutions seemingly cannot be beaten as the simplest ones! Congrats! Joachim

Generally, the all complex solutions of any equation (z - u)n = (z – w)n (n = 2, 3,...) with arbitrary fixed distinct complex numbers u = a + bi and w = c + di,are located on the line segment bisector of the segment |AB| with A(a, b) and B(c, d) (because of |z – u| = |z – w| is equivalent with d(z, A) = d(z, B)) (d(z, A) is the distance between the points z and A). Analytically, |z – u|2 = |z – w|2 is the line with equation z(\bar(w) - \bar(u)) + \bar(z)(w - u) + |u|2 - |w|2 = 0, or equivalently, Re(z(\bar(w) - \bar(u))) =(|w|2 - |u|2)/2. (\bar(z) denotes the conjugate complex number of z).

It seems worth to add, that continuing George Stoica

x_k = [ -1 + i tan (k \pi / n) ] / 2, for k=1,2,...,n-1.

Good job, George!

Dears

In addition to the nice answers, I think the simplest solution is as follows

|a+ib|n=|a+1+ib|n implies that |a+ib|=|a+1+ib|

then a2 +b2 = (a+1)2 + b2 hence, a = -1/2.

Best regards

Thank you, dear Issam for elaborating the details for what Romeo wrote briefly:

" the equality |x + 1|2 = |x|2 immediately yields Re(x) = -1/2 "

Dear Joachim Domsta

All answers are equivalent. Romeo goes with the geometric approach.

Mine is algebraic. Your answer shows a general solution. George attacks the problem through logical steps. Peter sheds light on some algebraic properties of the problem.

I think all answers provide a collection of nice different but equivalent solutions.

Best wishes