Real part of complex z must be zero 
Show that the real part of any solution of $(z+1)^{100}=(z-1)^{100}$
  must be zero.

I know that I must use the fact $e^{2\pi ki}$ somewhere but not sure exactly how or where to use it. 
 A: Take the modulus of each side: $|z+1|^{100} = |z-1|^{100}$.
Hence $|z+1| = |z-1|$ which means (interpret the modulus geometrically) that $z$ is on the perpendicular bisector of $-1$ and $1$.
A: Notice that the equation gives $|z+1|^{100} = |z-1|^{100}$, which gives $|z-1|=|z+1|$. Squaring gives $|z+1|^2 = |z|^2+z +\overline{z} +1 = |z|^2-z -\overline{z} +1 = |z-1|^2$. This gives $z +\overline{z}=0$, from which it follows that $\text{Re } z = 0$.
A: If $z=x+iy$ and if we put the value of $z$ in $|z+1|=|z-1|$ which has been derived in other answers,
$$|x+iy+1|=|x+iy-1|\implies (x+1)^2+y^2=(x-1)^2+y^2$$
$$\implies (x+1)^2-(x-1)^2=0\implies4\cdot x\cdot1=0\implies x=0$$
A little generalization:
Let $|z-w|=|z+w|$
If $z=x+iy,w=a+ib$ where $x,y,a,b$ are real and $a\cdot b\ne0$
$$(x-a)^2+(y-b)^2=(x+a)^2+(y+b)^2\implies x\cdot a+y\cdot b=0$$
If $b=0\implies w=a,x\cdot a=0\implies x=0$ as $a\ne0\implies z=iy$ (In our case, $a=1$)
If $a=0\implies w=ib,y\cdot b=0\implies y=0$ as $b\ne0\implies z=x$
Geometrically,  $$\arg(z)-\arg(w)=\arctan\frac yx-\arctan\frac ba$$
$$=\arctan \left(\frac{\frac yx-\frac ba}{1+\frac yx\cdot\frac ba}\right)=\arctan\left(\frac{ay-bx}{x\cdot a+y\cdot b}\right)=\frac\pi2 $$
$$\implies z\perp w $$
If $b=0,\arg(w)=\arctan 0=0$ or $\pi,$ so $w$ is parallel to $X$ axis, $z$ being perpendicular to $w,$ will be parallel to $Y$ axis. 
Observe that, $\arg(z)=\frac\pi2+0=\frac\pi2$ or $\frac\pi2+\pi\equiv-\frac\pi2$
Similarly  for $a=0$
