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Chapter 29: Problem 6

Es sei \(\zeta \neq 1\) eine \(n\)-te Einheitswurzel. Man zeige \(1+\zeta+\cdots+\zeta^{n-1}=0\).

Short Answer

Expert verified
The sum \(1+\zeta+\cdots+\zeta^{n-1}=0\) using the properties of a geometric series and \(\zeta^n=1\).

Step by step solution

01

Understand what is given

We are given that \(\zeta\) is an \(n\)-th root of unity but \(\zeta eq 1\). An \(n\)-th root of unity satisfies \(\zeta^n = 1\). We want to show that the sum of the first \(n\) powers of \(\zeta\), namely \(1 + \zeta + \zeta^2 + \ldots + \zeta^{n-1}\), is equal to 0.
02

Identify the sum formula

Recognize the sum \(1 + \zeta + \zeta^2 + \ldots + \zeta^{n-1}\) as a geometric series where the first term \(a = 1\) and the common ratio \(r = \zeta\). The number of terms in this series is \(n\).
03

Apply the geometric series sum formula

Apply the formula for the sum of a geometric series: \(S_n = a \frac{r^n - 1}{r - 1}\). Substituting \(a = 1\), \(r = \zeta\), and \(n\) as the number of terms, we get:\[S = \frac{\zeta^n - 1}{\zeta - 1}\]
04

Substitute \(\zeta^n = 1\)

Since \(\zeta\) is an \(n\)-th root of unity, we know that \(\zeta^n = 1\). Substitute this into the sum formula:\[S = \frac{1 - 1}{\zeta - 1} = \frac{0}{\zeta - 1} = 0\]
05

Conclusion

The calculation shows that the geometric series \(1 + \zeta + \zeta^2 + \ldots + \zeta^{n-1}\) sums to 0. Thus, it is confirmed that if \(\zeta\) is an \(n\)-th root of unity and \(\zeta eq 1\), then \(1 + \zeta + \zeta^2 + \ldots + \zeta^{n-1} = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This series has the form:
  • First term: \(a\)
  • Common ratio: \(r\)
  • Number of terms: \(n\)
The sum \((S_n)\) of a finite geometric series can be calculated using the formula:\[ S_n = a \frac{r^n - 1}{r - 1} \]This represents adding up all the terms of the series where:
  • \(a\) is the first term
  • \(r\) is the common ratio
  • \(n\) is the number of terms
In our example, the series \(1 + \zeta + \zeta^2 + \ldots + \zeta^{n-1}\) is a geometric series where:
  • The first term \(a = 1\)
  • The common ratio \(r = \zeta\)
  • The number of terms \(n\)
Understanding this series helps simplify calculating the sum of roots of unity, ultimately proving the sum equals zero.
Sum of Roots of Unity
Roots of unity are complex numbers that satisfy the equation \(z^n = 1\). The \(n\)-th roots of unity are the solutions to this equation. They are evenly spaced around the unit circle in the complex plane. These roots can be expressed in the form:\[ \zeta_k = e^{2\pi i k / n} \quad \text{for} \; k=0, 1, 2, \ldots, n-1 \]One specific property of these roots is that their sum is always zero. This can be analyzed through a geometric series:
  • If \(\zeta\) is an \(n\)-th root of unity, the series \(1 + \zeta + \zeta^2 + \ldots + \zeta^{n-1}\) indeed forms a complete cycle around the circle
  • This means adding all terms in the series results in a perfect balance, summing to zero
For \(\zetaeq 1\), the geometric series formula neatly shows us that:\[ S = \frac{1^n - 1}{\zeta - 1} = 0 \]Thus, this approach confirms the zero sum property of the sum of roots of unity in a simple, clear manner.
Complex Numbers
Complex numbers extend our traditional understanding of numbers, including real numbers and imaginary numbers. They have the form: \(z = a + bi\), where:
  • \(a\) is the real part
  • \(b\) is the imaginary part
  • \(i\) is the imaginary unit, satisfying \(i^2 = -1\)
In the context of roots of unity, complex numbers are crucial. They help us represent numbers that are not on the real number line and understand concepts like rotation around the unit circle. Each \(n\)-th root of unity corresponds to a point on this circle:
  • The unit circle is defined in the complex plane, centered at the origin with a radius of 1
  • It beautifully illustrates how these roots are spaced out evenly, providing a clear visual for reasoning about the sum of roots
Using complex numbers, we can efficiently calculate and derive properties like the sum of all roots being zero, as they naturally balance each other on the circle.

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Most popular questions from this chapter

Bestimmen Sie den Isomorphietyp der Galoisgruppe des Polynoms \(P=\) \(X^{4}+X^{3}+X^{2}+X+1 \in \mathbb{Z}_{2}[X]\) 眉ber \(\mathbb{Z}_{2}\)

F眉r welche \(n \in\\{1, \ldots, 100\\}\) ist ein regul盲res \(n\)-Eck konstruierbar?

Es seien \(\zeta_{1}, \ldots, \zeta_{n}\) die \(n\)-ten Einheitswurzeln. Man zeige \(\zeta_{1}^{k}+\zeta_{2}^{k}+\cdots+\zeta_{n}^{k}=0\) f眉r jedes \(k\) mit \(1 \leq k \leq n-1\).

Es gelte \(0

Es seien \(m, n\) nat眉rliche Zahlen mit \(\operatorname{ggT}(m, n)=1\) und \(\zeta_{m}\) eine primitive \(m\)-te Einheitswurzel 眉ber \(\mathbb{Q} .\) Zeigen Sie: Das \(n\)-te Kreisteilungspolynom \(\Phi_{n}(x) \in \mathbb{Q}[x]\) ist sogar 眉ber \(\mathbb{Q}\left(\zeta_{m}\right)\) irreduzibel.
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