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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.

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    $\begingroup$ Take $e^{i \pi / 3}$ and $e^{-i \pi / 3}$. This is the only solution. $\endgroup$ – Crostul Jan 4 '18 at 0:42
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    $\begingroup$ Suppose the root of unity is $x$. The other one must be $\bar x = x^{-1}$ since the imaginary part must cancel, so $x + x^{-1} = 1$ $\endgroup$ – Hw Chu Jan 4 '18 at 0:52
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    $\begingroup$ What you did is wrong. They need to be two different roots of unity $\endgroup$ – Dylan Jan 4 '18 at 1:37
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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.

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