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.
 A: HINT:
$\left(\dfrac{1+z}{1-z}\right)^8=1 \implies$
$\left(\dfrac{1+z}{1-z}\right)^4=\pm{1} \implies$
$\left(\dfrac{1+z}{1-z}\right)^2=\pm{1},\pm{i} \implies$
$\left(\dfrac{1+z}{1-z}\right)^1=\pm{1},\pm{i},\pm{\dfrac{1+i}{\sqrt{2}}},\pm{\dfrac{1-i}{\sqrt{2}}}$
A: You immediately get
$${1+z\over1-z}=\omega$$
where $\omega$ is any of the eighth roots of unity except, as it turns out, $-1$.  This solves in two steps:
$$1+z=\omega(1-z)\implies(1+\omega)z=\omega-1\implies z={\omega-1\over\omega+1}$$
(which explains why $\omega=-1$ is ruled out), and this gives
$$z={\omega-1\over\omega+1}\cdot{\overline\omega+1\over\overline\omega+1}={\omega-\overline\omega\over2+\omega+\overline\omega}={i\sin(\pi k/4)\over1+\cos(\pi k/4)}\quad\text{with}\quad-3\le k\le3$$
Plugging in the various $k$'s and simplifying, we get
$$z=0,\quad\pm i,\quad\pm(\sqrt2\pm1)i$$
This approach lends itself to arbitrary powers; there's nothing special here about $\omega$ being an eighth root of unity (aside from being able to evaluate the trig functions as radicals).
A: $\dfrac{1+z}{1-z}=e^{\frac{2ik\pi}{8}}$, for $k=1,\ldots,8$,  there is non-solution for $k=4$. You can find $z$ by solving the beginning equation  
A: 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
