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I actually have a follow-up question to this post -- given that n is a positive integer such that $z^n = (z+1)^n = 1$, I need to show that n is divisible by 6. I can now show that $z$ and $z+1$ both lie on the unit circle, but can this be of any help to me?

space
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    Draw a picture of two points on the unit circle that are exactly $1$ unit apart horizontally. What can you say about those points (in particular about their angle from the positive $x$-axis? – Steve Kass Nov 10 '17 at 15:48
  • Well I can see that the angle must be a 90 degrees angle or multiples of 90 – space Nov 10 '17 at 15:51
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    I don't think so. There are four points on the unit circle at multiples of $90^\circ$ from the positive $x$-axis, and only two of them are at the same horizontal position, but those two points $1+0i$ and $-1+0i$ are not one unit apart horizontally, so they can’t be $z$ and $z+1$. – Steve Kass Nov 10 '17 at 15:57
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    Oh so they must form an equilateral triangle with the center of the circle either facing down or up (not sideways)... – space Nov 10 '17 at 16:02
  • https://math.stackexchange.com/questions/643024/complex-numbers-exponential-numbers-proof – lab bhattacharjee Nov 10 '17 at 16:05

2 Answers2

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As you have noticed, $|z|=1$ and $|z+1|=1$ and so $z$ is in the intersection of the unit circle centered at the origin with the unit circle centered at $-1$.

The two intersection points of these circles are the primitive cubic roots of unit: $\omega$ and $\omega^2$.

Now $\omega^2+\omega+1=0$ and so $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$.

The result follows because $-\omega$ and $-\omega^2$ are primitive $6$-th roots of unit.

lhf
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Note that $$z^n=1 \implies z = e^{i2\pi k/n}, \tag 1$$ where $k \in \mathbb{Z}$. Likewise, $$(1+z)^n = 1 \implies 1+z = e^{i2\pi r/n} \implies 1=e^{i2\pi r/n}-e^{i2\pi k/n},\tag 2$$ where $r \in \mathbb{Z}$.

Now $$e^{i2\pi r/n}-e^{i2\pi k/n} = 2e^{i\pi (r+k)/n}\sin\left(\frac{\pi(r-k)}{n}\right)=1 \implies \sin\left(\frac{\pi(r-k)}{n}\right)=\pm \frac{1}{2}.\tag 3$$ Consequently, $$\frac{\pi(r-k)}{n}=\pm\frac{\pi}{6}+t\pi, \tag 4$$ where $t \in \mathbb{Z}$. Simplify $(6)$ to get $$n = \pm\frac{6(r-k)}{1+6t}.$$ Observe that $n$ is an integer, and $\gcd(6,1+6t)=1$. Therefore, $n=6p$, where $p \in \mathbb{Z}$.

Math Lover
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