$z^n=(z+1)^n=1$, show that $n$ is divisible by $6$.

Question

we are given $z^n=(z+1)^n=1$, $z$ complex number. we want to prove that $n$ is divisible by $6$. I showed that $|z|=|z+1|=1$. Hence $z$ is on the intersection of two unit circles, one centered at $(0,0)$ and the other one centered at $(-1,0)$. Then $z$ can take two values $z=\operatorname{cis}(2 \pi /3)$ or $z=\operatorname{cis}(4 \pi/3)$. Then from $z^n=1=\operatorname{cis}(2 \pi k)$, $k$ integer I showed that $n$ should be divisible by $3$. I am left to show that $n$ should be divisible by $2$. I think if I figure out a way to show that $\operatorname{arg}(z+1)=\pi/3$ or $-\pi/3$ then I would be done, by using $(z+1)^n=\operatorname{cis}(2\pi k)$. but how can I do this? Any ideas are welcome! Thanks in advance!

Answer 1

If $z$ is one of the two nontrivial roots of unity, then $z+1$ is...?

Geometrically: Consider the hexagon inscribed inside the unit circle.

Algebraically: if $z^3=1$ then $z^2+z+1=0$ so $z+1=-z^2$ is a $6$th root.

Answer 2

If $z=\operatorname{cis}\left(\dfrac{2\pi}3\right)$ then what is $z+1$?

It is \begin{align} & \left(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\right)+1 \\[6pt] = {} & \left(-\frac 1 2 + i\frac{\sqrt 3}2\right)+1 \\[6pt] = {} & \frac 1 2 + i \frac{\sqrt 3} 2 \\[6pt] = {} & \operatorname{cis}\left( \frac \pi 3 \right). \end{align}

It's not hard to see that if you raise that to the $6$th power you get $1$ but to any lower power you get something other than $1$.

Answer 3

If you write your two possible values of $z$ as $\frac{-1}{2} \pm \frac{i\sqrt{3}}{2}$, then you are asking to find the argument of $z+1 = \frac{1}{2} \pm \frac{i\sqrt{3}}{2}$, which are standard values.

Answer 4

$z^n=1\implies z=\cos\tfrac{2\pi a}n+i\sin\tfrac{2\pi a}n$ where $a\equiv0,1,\cdots,n-1\pmod n$

$(z+1)^n=1\implies z+1=\cos\tfrac{2\pi b}n+i\sin\tfrac{2\pi b}n$ where $b\equiv0,1,\cdots,n-1\pmod n$

Equating the imaginary parts, $\sin\tfrac{2\pi a}n=\sin\tfrac{2\pi b}n$

$\implies\cos\tfrac{2\pi a}n=\pm\cos\tfrac{2\pi b}n$

Equating the real parts, $\cos\tfrac{2\pi a}n=\cos\tfrac{2\pi b}n-1$

Clearly, $\cos\tfrac{2\pi a}n=\cos\tfrac{2\pi b}n$ gives absurdity

$\implies\cos\tfrac{2\pi a}n=-\cos\tfrac{2\pi b}n$

Consequently, $-\cos\tfrac{2\pi b}n=\cos\tfrac{2\pi b}n-1\iff\cos\tfrac{2\pi b}n=\tfrac12=\cos\tfrac\pi3$

$\implies\tfrac{2\pi a}n=2m\pi\pm\tfrac\pi3=\tfrac\pi3(6m\pm1)$

$\iff\dfrac{n(6m\pm1)}6=a$ which is an integer

$\implies6|n(6m\pm1)$

But $(6m\pm1,6)=1$