If $z$ is a complex number and $z^n=1$, then $z = \pm1$ and therefore $|z| = 1$. Then similarly, $(z+1)^n = 1$ so $(z+1) = \pm1$, which also implies that $|z+1| = 1$.
So I'm writing a mathematical proof for a problem, and I was just wondering if my saying $z=\pm1$ and the like is right because I seem to remember something about the $\pm$ not being valid in the complex numbers. If so, given $z^n = (z + 1)^n = 1$, how can I prove that $|z| = 1$ and $|z+1| = 1$?
If $z^n=1$, and $n\in \mathbb{N}$, then $z=e^{i2k\pi/n}$, $k=1,\dots,n$ are the $n$ roots. Since $|e^{i\phi}|=1$ for any real valued $\phi$, then $z^n=1\implies |z|=1$.
Note that for $k=n$, $z=1$; so, $z=1$ is a root.
If $n$ is even, then clearly $z=-1$ is a root ($k=n/2$), while if $n$ is odd, then $z=-1$ is not a root.
If $(z+1)^n$, then the previous analysis holds with $z$ replaced by $z+1$ .
As also mentioned in other answers, $z^n=1$ does not imply that $z=\pm 1$ in general.
Regarding the proof of the absolute values, we use the following property of the absolute values of two complex numbers. $$|ab|=|a||b|. \tag 1$$ Consequently, $$|z^n|=|z|^n=1 \implies |z|=1, \tag 2$$ where the implication uses the fact that $|z|$ cannot be negative.
Likewise, $(z+1)^n=1$ implies $|z+1|=1$.
