Complex numbers - Exponential numbers - Proof

Question

Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.

For this problem I am stumped...how should I begin? Also there's a hint for it:

From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z + 1)^n = 1$ tell you? What do the resulting equations tell you about $z$?

Could someone give me a hint on where to begin? thanks in advance

Answer 1

Hint 1: $|z-z_0|=r$ is a circle with radius $r$ and center $z_0$;

Hint 2: the unit circle is $z=e^{i\theta}$;

Hint 3: for any $n\in\Bbb{N}$, $(e^{i\theta})^n=e^{in\theta}$;

Hint 4: if $\cos \theta= \pm 1$ and $\sin \theta =0$ then $\theta=k\pi$ for some integer $k$;

Hint 5: if $a,b,c$ are integers such that $a|bc$ and $\gcd(a,b)=1$ then $a|c$.

Answer 2

A problem way too cool and cute to pass up, so check this out:

$(1.) \; z^n = 1 \Rightarrow \vert z \vert^n = 1, \tag{1}$

$(2.) \; \vert z \vert^n = 1 \Rightarrow \vert z \vert = 1 \Rightarrow \exists \theta \in \Bbb R \;\text{such that} \; z = e^{i\theta}, \tag{2}$

$(3.) \; \vert z \vert = 1 \Rightarrow z \bar z = 1, \tag{3}$

$(4.) \; (z + 1)^n = 1 \Rightarrow \vert z + 1 \vert^n = 1 \Rightarrow \vert z + 1\vert = 1, \tag{4}$

$(5.) \; \vert z + 1 \vert = 1 \Rightarrow (1 + z)(1 + \bar z) = 1, \tag{5}$

$(6.) \; (1 + z)(1 + \bar z) = 1 \Rightarrow z \bar z + z + \bar z + 1 = 1 \Rightarrow z \bar z + z + \bar z = 0, \tag{6}$

$(7.) \; \text{by (3) and (6),} \; z + \bar z = -1, \tag{7}$

$(8.) \; \text{by (2) and (7),} \; 2 \Re{z} = 2 \cos \theta = -1 \Rightarrow \cos \theta = -\dfrac{1}{2}, \tag{8}$

$(9.) \; \cos \theta = -\dfrac{1}{2} \Rightarrow \Re{(1 + z)} = 1 + \cos \theta = \dfrac{1}{2}, \tag{9}$

$(10.) \; \Re{(1 + z)} = \dfrac{1}{2} \; \text{and} \; \vert 1 + z \vert = 1 \Rightarrow 1 + z = e^{(\pm 2\pi i/ 6) + 2k \pi}, k \in \Bbb Z, \tag{10}$

$(11.) \; 1 + z = e^{(\pm 2\pi i/ 6) + 2k \pi} \Rightarrow 1 = (1 + z)^n = e^{\pm 2n \pi i / 6}, \tag{11}$

$(12.) \; e^{\pm 2n \pi i / 6} = 1 \Rightarrow 6 \mid n. \tag{12}$

QED

Answer 3

We assume $r,m,v$ to be integers

Using de Moivre's theorem, $$z^n=1=e^{2r\pi i}\implies z=e^{\dfrac{2r\pi i}n}=\cos\frac{2r\pi}n+i\sin \dfrac{2r\pi}n$$

From this, $$z^n=(1+z)^n\implies z=-\frac12-i\frac{\cot\dfrac{m\pi}n}2$$

So, equating the real parts, $\displaystyle\cos\frac{2r\pi}n=-\frac12$

$\displaystyle\implies\sin\dfrac{2r\pi}n=\pm\sqrt{1-\left(-\frac12\right)^2}=\pm\frac{\sqrt3}2$

Equating the imaginary part,

$$-\frac{\cot\dfrac{m\pi}n}2=\pm\frac{\sqrt3}2\implies\tan\dfrac{m\pi}n=\mp\frac1{\sqrt3}=\tan\left(\mp\frac\pi6\right)$$

$$\implies \dfrac{m\pi}n=v\pi\mp\frac\pi6\iff 6m=n(6v\mp1)$$

We have $\displaystyle\frac{n(6v\mp1)}6=m$ which is an integer

But as $\displaystyle(6v\mp1,6)=1, 6$ must divide $n$

Answer 4

We can say: $$z^n=1 \implies |z|^n=1 \implies |z^n|=1 \implies |z|=1$$ Using similar logic, we can also get: $$(z+1)^n=1 \implies |z+1|^n=1 \implies |(z+1)^n|=1 \implies |z+1|=1$$ Let's set $z=a+bi$: $$|z|=|z+1| \implies a^2 + b^2 = (a+1)^2 + b^2 \implies a^2 + b^2 = a^2 + b^2 + 2a + 1$$ $$\implies 2a + 1 = 0 \implies a = -\frac{1}{2}$$ Pluging this back into the equation, we get: $$\left(-\frac{1}{2}\right)^2 + b^2 = 1 \implies b^2 = \frac{3}{4} \implies b = \pm \frac{\sqrt{3}}{2}$$ With this information, we know that: $$z= -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \implies z = e^{\pm 2\pi i/3}$$ Also: $$z+1= \frac{1}{2} \pm \frac{\sqrt{3}}{2}i \implies z+1 = e^{\pm \pi i/3} \implies (z+1)^n = e^{\pm \pi ni/3}$$ For $(z+1)^n$ to be $1$, $\cos \pm \pi n/3 = 1$ and $\sin \pm \pi n/3 = 0$. This can only happen if $\pm \pi n/3 = 2\pi k$ for some integer $k$ Solving for $n$, we get: $$\pm \pi n/3 = 2\pi k \implies n = \pm 6k \implies \boxed{6 \mid n}.$$

Answer 5

We assume $u,v,m,p $ to be integers

Using de Moivre's theorem and Euler Formula, $$z^n=1=e^{2u\pi i}\implies z=e^{\dfrac{2u\pi i}n}=\cos\frac{2u\pi}n+i\sin \dfrac{2u\pi}n$$

$$\text{and }(z+1)^n=1\implies z+1=\cos\frac{2v\pi}n+i\sin\dfrac{2v\pi}n$$

$$\implies \cos\frac{2u\pi}n+i\sin\dfrac{2u\pi}n=\cos\frac{2v\pi}n-1+i\sin\dfrac{2v\pi}n$$

Equating the real parts, $\displaystyle\cos\dfrac{2u\pi}n=\cos\dfrac{2v\pi}n-1\ \ \ \ (1)$

Again, equating the imaginary parts, $\displaystyle\sin\dfrac{2u\pi}n=\sin\dfrac{2v\pi}n \ \ \ \ (2)$

$\displaystyle\implies\cos\dfrac{2u\pi}n=\pm\cos\dfrac{2v\pi}n$

Taking the '+' sign, $(1)$ reduces to $\displaystyle\cos\dfrac{2v\pi}n=\cos\dfrac{2v\pi}n-1\iff 0=-1$

Taking the '-' sign, $(1)$ reduces to $\displaystyle-\cos\dfrac{2v\pi}n=\cos\dfrac{2v\pi}n-1\iff\cos\dfrac{2v\pi}n=\frac12=\cos\frac\pi3$

$\displaystyle\implies\dfrac{2v\pi}n=2p\pi\pm \frac\pi3\implies \frac{n(6p\pm1)}6=v$ which is an integer (Now, the last logic is omitted as is available from my other answer)

One observation:

$z$ is actually $\omega$ where $\omega$ is a Complex Cube root of Unity

Also try squaring & adding $(1),(2)$