Adding complex exponentials

Question

Can somebody please explain

$$e^{-\frac{3}{4}\pi i}+e^{-\frac{9}{4}\pi i}+e^{-\frac{15}{4}\pi i}+e^{-\frac{21}{4}\pi i}=0$$ WolframAlpha Computation.

Answer 1

I'll explain in two ways - the first more intuitively.

First method: If you think of the sum as adding vectors in the complex plane, you can easily see that the sum is $0$.

$$\color{green}{e^{-\frac{3}{4}\pi i}}+\color{red}{e^{-\frac{9}{4}\pi i}}+\color{pink}{e^{-\frac{15}{4}\pi i}}+\color{purple}{e^{-\frac{21}{4}\pi i}}=0$$

Second method: Algebraicly, we can use the fact that $$-e^{i\theta}=e^{i\theta+\pi}=e^{i\theta+3\pi}$$ Now if we choose $\theta_1=-\frac{15}{4}\pi$ and $\theta_2=-\frac{21}{4}\pi$, we have $$e^{-\frac{3}{4}\pi i}+e^{-\frac{9}{4}\pi i}+e^{-\frac{15}{4}\pi i}+e^{-\frac{21}{4}\pi i}=-e^{-\frac{15}{4}\pi i}-e^{-\frac{21}{4}\pi i}+e^{-\frac{15}{4}\pi i}+e^{-\frac{21}{4}\pi i}=0$$

Answer 2

Remember that $e^{i\theta} = \cos\theta+ i\sin\theta$ and so $e^{2\pi i} = 1$. So you can mark these four points on the unit circle, and you'll see that they are at angles, respectively, $-3\pi/4$, $-\pi/4$, $\pi/4$, and $3\pi/4$. Thus, these are the vertices of a square, and the four complex numbers add up to $0$.

Answer 3

Another approach can be seen in the following: \begin{align*} e^{-3\pi i/4} + e^{-i\pi i/4} + e^{-15\pi i/4} + e^{-21\pi i/4} & = e^{5\pi i/4} + e^{7\pi i/4} + e^{\pi i/4} + e^{3\pi i/4} \\ & = e^{\pi i/4}\left(e^{\pi i} + e^{3\pi i/2} + 1 + e^{\pi i/2} \right).\end{align*} Now, the sum \begin{equation} e^{\pi i} + e^{3\pi i/2} + 1 + e^{\pi i/2}\end{equation} is of the four distinct roots of unity. That is, each of the above satisfies the polynomial $$x^4 - 1 = 0.$$ Let's write $e^{\pi i/2} = \omega$. Then the sum in question can be rewritten to read $$e^{\pi i} + e^{3\pi i/2} + 1 + e^{\pi i /2} = 1 + e^{\pi i/2} + e^{\pi i} + e^{3\pi i/2} = \omega^3 + \omega^2 + \omega + 1.$$ Notice that $$0 = \omega^4 - 1 = (\omega - 1)\left(\omega^3 + \omega^2 + \omega + 1\right)$$ and since $e^{\pi i/4} \ne 1$, we have that it must be the case that $$\omega^3 + \omega^2 + \omega + 1 = 0.$$ The result follows.

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It is worth noting that this argument can be generalized to other settings involving higher roots of unity and demonstrates why ``the sum of vertices of a square on the unit circle equals 0".