Let $w \to w^{a}$ be the principal branch of the power function defined for $|\mathrm{Arg}(w)| <\pi$. Find the set of all values of $z\in \mathbb{C}$ such that the following identity holds for ALL $a\in \mathbb{R}$.
$$(z^2-1)^{a}=(z+1)^a(z-1)^a$$
My approach: Based on the definition for complex exponent we have $$\mathrm{exp}(a\mathrm{Log}(z^2-1))=\mathrm{exp}(a\mathrm{Log}(z-1))\mathrm{exp}(a\mathrm{Log}(z+1))$$ Which is equivalent to $$a\left[\mathrm{Arg}((z+1)(z-1))-(\mathrm{Arg}(z+1)+\mathrm{Arg}(z-1))\right]\in2\pi\mathbb{Z}$$
Since it holds for all $a\in \mathbb{R}$, then $$\mathrm{Arg}((z+1)(z-1))-(\mathrm{Arg}(z+1)+\mathrm{Arg}(z-1))=0$$
A quick sketch on paper (yet to be rigorously computed) shows that $z$ may take any value on the real set: $\mathbb{R}-\left(\{1\}\cup (-\infty,-1]\right)$. $z$ may also take any value on the positive part of the imaginary axis ($y$-axis) and may lie in the first and the fourth quadrant too.
However, $z$ can NOT take any value in the second or third quadrants or on the negative part of the $y$ axis.
Any comments about my approach and conclusions?
If you take $$ \mathrm {Arg}(w)\in (-\pi,\pi] $$ then everything will be ok..