Complex numbers: proof of equality

Question

Can anyone help me prove following equality?

$$\left(\frac{-1+i\sqrt{3}}{2}\right)^n + \left(\frac{-1-i\sqrt{3}}{2}\right)^n$$ $$=2 \iff \frac{n}3\in \mathbb{N}$$ $$=-1 \iff \frac{n}3\not\in \mathbb{N}$$ This is what I've got: $$\left(\frac{-1+i\sqrt{3}}{2}\right)^n + \left(\frac{-1-i\sqrt{3}}{2}\right)^n$$ $$= \left(\frac{2(cos(-\pi/3)+i sin(-\pi/3))}2\right)^n + \left(\frac{2(cos(\pi/3)+i sin(\pi/3))}2\right)^n$$ $$=[cos(-\pi/3)+isin(-\pi/3)]^n-[cos(\pi/3)+isin(\pi/3)]^n$$ $$=cos(n\pi/3)-isin(n\pi/3)-cos(n\pi/3)-isin(n\pi/3)$$ $$=-2isin(n\pi/3)$$ and if $n/3 \in \mathbb{N}$, I get $n'\pi$ and $sin(n'\pi)=0$ which isn't the result I needed...

Answer 1

Lets write $$\zeta=\frac{-1+i\sqrt{3}}{2}$$

then $$\zeta^{-1}=\zeta^2=\frac{-1-i\sqrt{3}}{2}$$

Now $\zeta$ is a cube root of $1$, so $\zeta^3=1$ and also $\zeta^{-3}=1$

Thus if $3$ divides $n$ then

$$\zeta^n+\zeta^{-n}=1+1=2$$

If $3$ does not divide $n$ then $n\equiv 1 \ \text{or} \ -1 \mod 3$ and so

$$\zeta^n+\zeta^{-n}=\zeta+\zeta^{-1}=-1$$

Answer 2

Since $$ \left(\frac{-1+i\sqrt{3}}{2}\right)^3=1=\left(\frac{-1-i\sqrt{3}}{2}\right)^3 $$ you know that both numbers are roots of the polynomial $x^2+x+1$.

For $n=0$ we get $2$; for $n=1$ we get $-1$. So the given expression is the solution of the recurrence $$ a_{n+2}=-a_{n+1}-a_n,\quad a_0=2, \quad a_1=-1 $$ We have $$ a_{n+3}=-a_{n+2}-a_{n+1}=a_{n+1}+a_n-a_{n+1}=a_n $$

Answer 3

Note that using Euler's Formula, we can write $\frac{-1\pm i\sqrt3}{2}=\cos(2\pi/3)\pm i\sin(2\pi/3)=e^{\pm i2\pi/3}$.

Therefore, from De Moivre's Formula, we have

$$\begin{align} \left(\frac{-1+i\sqrt3}{2}\right)^n+\left(\frac{-1-i\sqrt3}{2}\right)^n&=e^{i2n\pi/3}+e^{-i2n\pi/3}\\\\ &=2\cos(2n\pi/3) \end{align}$$

Finally, we see that

$$\left(\frac{-1+i\sqrt3}{2}\right)^n+\left(\frac{-1-i\sqrt3}{2}\right)^n=\begin{cases}2&,n=0,3,6\dots\\\\-1&,n\ne0,3,6\dots\end{cases}$$

Answer 4

The idea is good, but your angles are not quite right: for example, $\cos{(-\pi/3)} = \cos{(\pi/3)} = 1/2$ and you want an angle $\alpha$ with $\cos{\alpha} = -1/2$ and $\sin{\alpha} = \sqrt{3}/2$. One such angle is $\alpha = 2\pi/3$, leading to $2\cos{(2n\pi/3)}$, which is equal to 2 if and only if $n/3$ is an integer and to $-1$ otherwise. [Note that the question also makes sense for negative integers $n$.]

Answer 5

Theres no point in converting to trig unless you are familiar and comfortable with polar coordinates. If so:

$(\frac {-1 + i\sqrt{3}}2)^n+(\frac {-1 - i\sqrt{3}}2)^n=$

$(\cos \frac {2\pi}{3} + i \sin \frac {2\pi}{3})^n + (\cos \frac {4\pi}{3} + i \sin \frac {4\pi}{3})^n=$

$(e^{i\frac{2\pi}{3}})^n + (e^{i\frac{4\pi}{3}})^n=e^{i\frac{2n\pi}{3}} + e^{i\frac{4n\pi}{3}}=$

$(\cos \frac {2n\pi}{3} + i \sin \frac {2n\pi}{3}) + (\cos \frac {4n\pi}{3} + i \sin \frac {4n\pi}{3})=$

$3|n$ i.e. if $n = 3k$

$= \cos 2k\pi + i \sin 2k \pi + \cos 4k\pi + i\sin 4k\pi = 1 + 0i + 1 + 0i = 2$

if $3\not \mid n$ ie. if $n = 3k \pm 1$

$ = \cos (\pm \frac{2\pi}3) + i \sin (\pm \frac{2\pi}3) + \cos (\pm \frac{4\pi}3) + i \sin (\pm \frac{4\pi}3)$

$= -\frac 12 \pm i\frac{\sqrt{3}}{2} -\frac 12 \mp i \frac{\sqrt{3}}{2}= -1$

But really itd be easier to realize: $(\frac {-1 \pm i\sqrt{3}}2)^3 = 1$ and $(\frac {-1 \pm i\sqrt{3}}2)^2 = \frac {-1 \mp i\sqrt{3}}2$ [just multiply them out]so

$(\frac {-1 + i\sqrt{3}}2)^{3k + i} + (\frac {-1 - i\sqrt{3}}2)^{3k + i} = $

$(\frac {-1 + i\sqrt{3}}2)^i + (\frac {-1 - i\sqrt{3}}2)^i = 1 + 1 = 2$ if $i = 0$

$=(\frac {-1 \pm i\sqrt{3}}2) + (\frac {-1 \mp i\sqrt{3}}2)$

$= -\frac 12 - \frac 12 =-1$ if $i = 1$ or $i = 2$.