TITLE: Show that if $(z+1)^{100} = (z-1)^{100}$, then $z$ is purely imaginary
QUESTION [8 upvotes]: Let $z$ be a complex number satisfying
  $$(z+1)^{100} = (z-1)^{100}$$
  Show that $z$ is purely imaginary, i.e. that $\Re(z) = 0$.

Rearrange to 
$$\left(\frac{z+1}{z-1}\right)^{100} = 1$$
I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^{i\pi}$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!

REPLY [6 votes]: Note that $|z+1|=|z-1|$ 

Let $z=x+iy$
Therefore 
$\sqrt {(x+1)^2+y^2}=\sqrt {(x-1)^2+y^2}$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$ 
$4x=0$ 
$\therefore x=0$
As we have found out, $z=x+iy\implies z=iy$ therefore yes, it is purely imaginary.