<s>[INST] 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.
 [/INST] HINT:
$\left(\dfrac{1+z}{1-z}\right)^8=1 \implies$
$\left(\dfrac{1+z}{1-z}\right)^4=\pm{1} \implies$
$\left(\dfrac{1+z}{1-z}\right)^2=\pm{1},\pm{i} \implies$
$\left(\dfrac{1+z}{1-z}\right)^1=\pm{1},\pm{i},\pm{\dfrac{1+i}{\sqrt{2}}},\pm{\dfrac{1-i}{\sqrt{2}}}$
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