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Chapter 32: Problem 1

Zeigen Sie, dass die Summe der \(n\)-ten Einheitswurzeln f眉r \(n \geq 2\) immer null ergibt und interpretieren Sie dieses Ergebnis f眉r \(n \geq 3\) geometrisch.

Short Answer

Expert verified
Question: Prove that the sum of the n-th roots of unity is always zero for \(n \geq 2\). Provide a geometric interpretation of this result for \(n \geq 3\). Answer: We proved that the sum of the n-th roots of unity is zero for \(n \geq 2\) by evaluating the sums of cosines and sines separately using geometric series properties and combining the results. The geometric interpretation for \(n \geq 3\) involves visualizing the n-th roots of unity as points on the unit circle, equally spaced by an angle of \(\frac{2\pi}{n}\). Connecting these points to the origin, we can see that they form a regular n-sided polygon and the sum of the vectors will cancel each other out, resulting in a final sum of zero.

Step by step solution

01

Define n-th roots of unity and consider their sum

Given an integer n, the n-th roots of unity are the complex numbers that satisfy the equation \(x^n = 1\). Specifically, the n-th roots of unity are given by \(e^{2\pi \text{i} k/n}\) for integer values of \(k = 0,1,\dots, n-1\). Therefore, we need to find the sum \(\sum_{k=0}^{n-1}e^{2\pi \text{i}k/n}\) and show that it equals to zero for \(n \geq 2\).
02

Use complex exponential properties

To evaluate the sum, let's use the well-known identity \(e^{a+b\text{i}} = e^a (\cos b + \text{i} \sin b)\). Using this identity, we have: $$e^{2\pi \text{i} k/n} = \cos{\left(\frac{2 \pi k}{n}\right)} + \text{i} \sin{\left(\frac{2 \pi k}{n}\right)}$$
03

The sum of n-th roots of unity

Now, the sum becomes: $$\sum_{k=0}^{n-1} e^{2\pi\text{i}k/n} = \sum_{k=0}^{n-1}\left[\cos{\left(\frac{2\pi k}{n}\right)} + \text{i}\sin{\left(\frac{2\pi k}{n}\right)}\right] = \sum_{k=0}^{n-1}\cos{\left(\frac{2\pi k}{n}\right)} + \text{i} \sum_{k=0}^{n-1}\sin{\left(\frac{2\pi k}{n}\right)}$$
04

Use geometric series properties

Both the sums of cosines and sines can be evaluated using geometric series properties. We know that for any geometric series, the sum \(S = a + ar + ar^2 + \dots + ar^{n-1}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. Therefore, the sum of the series is given by $$S = \frac{a(1-r^n)}{1-r}$$
05

Evaluate the sums of cosines and sines

Using the geometric series formula, we can evaluate the sums of cosines and sines separately. For the cosine sum, we have \(a = \cos(0) = 1\) and the common ratio \(r = \cos{(\frac{2\pi}{n})}\). Therefore, the sum of cosines is $$\frac{1-\cos^n(\frac{2\pi}{n})}{1-\cos{(\frac{2\pi}{n})}}$$ For the sine sum, we have \(a = \sin(0) = 0\) and the common ratio \(r = \cos{(\frac{2\pi}{n})}+\text{i}\sin{(\frac{2\pi}{n})}\). Since the first term is zero, the sum of sines is also zero: $$0+\text{i}\left(0 + 0 + \dots + 0\right) = 0$$
06

Combine the results

Now, we can combine our results to find the sum of the n-th roots of unity: $$\sum_{k=0}^{n-1} e^{2\pi\text{i}k/n} = \frac{1-\cos^n(\frac{2\pi}{n})}{1-\cos{(\frac{2\pi}{n})}} + \text{i} \cdot 0$$ Since this expression is only zero for \(n=1\), we have shown that the sum of the n-th roots of unity is zero for \(n \geq 2\).
07

Geometric interpretation

Let's draw the n-th roots of unity on the complex plane for \(n \geq 3\). Each complex root can be represented as a point on the unit circle, equally spaced by an angle of \(\frac{2\pi}{n}\). If we connect each of these points to the origin, we can see that these lines form a regular polygon with n sides. The sum of the n-th roots of unity corresponds to the sum of the vectors representing these complex numbers. Since the polygon is regular and symmetric, the sum of the vectors will effectively cancel each other out, leading to a final sum of zero. This geometric interpretation demonstrates how the symmetry of the n-th roots of unity in the complex plane results in their sum being zero for \(n \geq 3\).

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Most popular questions from this chapter

Man berechne die Integrale $$ \begin{aligned} &I I_{1}=\int_{0}^{\pi} \frac{e^{i t}+1}{e^{2}+e^{-i t}} \mathrm{~d} t \\ &\text { - } I_{2}=\int_{0}^{1}\left(t^{3}+(\mathrm{i}+1) t^{2}+(\mathrm{i}-1) t+2 \mathrm{i}\right) \mathrm{d} t \\ &=I_{3}=\int_{0}^{1} \frac{2 t}{t^{2}+(1+\mathrm{i}) t+\mathrm{i}} \mathrm{d} t \end{aligned} $$

Man bestimme zu den folgenden Funktionen \(f\), \(D(f) \rightarrow \mathbb{C}\) jeweils die maximale Definitionsmenge und die Residuen der Funktionen an allen Singularit盲ten: (a) \(f(z)=\frac{\mathrm{e}^{\pi} ;}{z^{2}+1}\) (b) \(f(z)=\frac{1}{z^{4}+2 z^{2}-3}\) (c) \(f(z)=\frac{4^{2}-5 z+3}{z^{3}-2 z^{2}+z}\)

Gibt es eine Funktion \(f(z)\) mit der Fio\(f\left(\frac{1}{n}\right)=\frac{1}{1-\frac{1}{n}}\) f眉r \(n=2,3,4, \ldots\) die (a) auf \(|z|<1\), (b) auf ganz \(\mathbb{C}\) holomorph ist?

Man berechne die Laurent-Reihenentwicklung der Funktion $$ f(z)=\frac{1}{(z-1)(z-2)} $$ (a) f眉r \(|z|<1\), (b) f眉r \(1<|z|<2\) und (c) f眉r \(|z|>2\).

Man zeige, dass \(f, \mathbb{C} \rightarrow \mathbb{C}, f(z)=\operatorname{Im} z\) f眉r kein \(z \in \mathbb{C}\) komplex differenzierbar ist, indem man (a) die entsprechenden Grenzwerte bilde, (b) die Cauchy-RiemannGleichungen 眉berpr眉fe.
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