Solve $(1+z)^8=(1-z)^8$ My guess is to write this as $$\left(\frac{1+z}{1-z}\right)^8=1.$$ We can then find 8 possibilities for $\frac{1+z}{1-z}$, namely $\cos(k\pi/4)+i\sin(k\pi/4)$, $k=1,\ldots,8$. For each $k$ we can then deduce 2 equations by putting $z=x+iy$, for example for $k=1$ we get: $$\frac{1+x+iy}{1-x-iy}=\frac12 (1+i).$$ Now we find two equations for $x$ and $y$, by noting both the real and imaginary parts of the equations should be equal, and can thus find $z$.
However, doing this for $k=1,\ldots,8$ seems somewhat cumbersome. Does anyone know of a faster way to find the $z$? Thans in advance.

This reduces to:
$$1+8x+28x^2+56x^3+70x^4+56x^5+28x^6+8x^7+x^8=1-8x+28x^2-56x^3+70x^4-56x^5+28x^6-8x^7+x^8$$Which becomes $$16x+112x^3+112x^5+16x^7=0$$
Whence $x=0$ or $$x^6+7x^4+7x^2+1=0$$
Write $y=x^2$ to obtain $$y^3+7y^2+7y+1=0$$
We have the obvious solution $y=-1$ and $$y^2+6y+1=0$$ for the others
