# Nth roots of unity

Find two roots of unity such that their sum equal 1 Okay so what I did, is that I said that: $$\exp((2i\pi)/n)+\exp((2i\pi)/n)=1$$ which was equivalent to $$2\exp((2i\pi)/n)=1$$

$$\exp((2i\pi)/n)=\frac 1 2$$ But then only non-sense came from this I don't know why.

• Take $e^{i \pi / 3}$ and $e^{-i \pi / 3}$. This is the only solution. Commented Jan 4, 2018 at 0:42
• Suppose the root of unity is $x$. The other one must be $\bar x = x^{-1}$ since the imaginary part must cancel, so $x + x^{-1} = 1$ Commented Jan 4, 2018 at 0:52
• What you did is wrong. They need to be two different roots of unity Commented Jan 4, 2018 at 1:37

If $w$ is a root of unity then $w^n=1\implies |w|=1$. So $w$ belongs to the unit circle of radius $1$ centred in $(0,0)$.

Let assume $1-w$ is also a root of unity, then $|1-w|=1$ and $w$ also belongs to the circle of radius $1$ centred in $(1,0)$.

The intersection of these two circles have abscissa $\dfrac 12$ and correspond to the angle $\dfrac{\pi}3$.

[since $\cos(\frac\pi3)=\frac 12$].

And you can verify that $w=e^{i\pi/3}=\dfrac{1+i\sqrt{3}}2$ and $1-w=e^{-i\pi/3}=\dfrac{1-i\sqrt{3}}2$

both verify $w^6=1$ and have sum $1$.

Regarding your approach you should solve instead $e^{i2k_1\pi/n}+e^{i2k_2\pi/n}=1$.

Here you assumed wrongly $k_1=k_2=1$.

By developping this we get $\begin{cases}\cos(\frac{2k_1\pi}n)+\cos(\frac{2k_2\pi}n)=1\\\sin(\frac{2k_1\pi}n)+\sin(\frac{2k_2\pi}n)=0\end{cases}$

The second equation forces $\theta_1=\theta_2+\pi$ or $\theta_1=-\theta_2$

In the first case, the cosinus sum also vanish, thus we have $2\cos(\frac{2k_1\pi}n)=1\iff \dfrac{2k_1\pi}n=\pm\dfrac{\pi}3$

And we get the same solution than previously.

Having derived $$2\Re[\exp(i2\pi/n)]=1$$, proceed as follows, starting with Euler's Formula:

$$2\cos(2\pi/n)=1$$

$$2\cos(2\pi/n)\sin(2\pi/n)=\sin(2\pi/n)$$

$$\sin[2(2\pi/n)]=\sin(\pi/2n)$$ (from the double-angle formula for sines or the sum-product relation for $$\cos u\sin v$$)

So $$2(2\pi/n)-2\pi/n=2\pi/n$$ is an even number times $$\pi$$ or else $$2(2\pi/n)+2\pi/n=6\pi/n$$ is an odd number times $$\pi$$. The only whole numbers for $$n$$ which satisfy at least one of these relations are $$1,2,6$$, and just one of those three candidates actually survives checking.