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Sei \(n \in \mathrm{N}\) und \(W(n)=\left\\{z \in \mathbb{C} ; z^{n}=1\right\\}\) die Menge der \(n\)-ten Einheitswurzeln. Man zeige: a) \(W(n)\) ist cine Untergruppe von \(C^{*}\) (und damit selbst eine Gruppe). b) \(W(n)\) ist eine zyklische Gruppe der Ordnung \(n, d . h\). es gibt ein \(\zeta \in W(n)\) mit $$ W(n)=\left\\{\zeta^{*} ; \quad 0 \leq u

Short Answer

Expert verified
W(n) is a cyclic subgroup of \( \mathbb{C}^* \) with order \( n \) and \( \varphi(n) \) primitive roots.

Step by step solution

01

Definition of Unit Roots

The set \( W(n) = \{ z \in \mathbb{C} \, ; \, z^n = 1 \} \) represents the set of \( n \)-th unit roots, which are complex numbers whose \( n \)-th power equal 1.
02

Show W(n) is a Subgroup of C*

To prove that \( W(n) \) is a subgroup of \( \mathbb{C}^* \), we need to show three things: non-empty, closed under multiplication, and closed under taking inverses. \( W(n) \) contains the identity, 1, because \( 1^n = 1 \). If \( z_1, z_2 \in W(n) \), then their product \( z_1 z_2 \) is also in \( W(n) \) since \( (z_1 z_2)^n = z_1^n z_2^n = 1 \cdot 1 = 1 \). Lastly, the inverse \( z^{-1} \) is in \( W(n) \) as well because \( (z^{-1})^n = (z^n)^{-1} = 1^{-1} = 1 \).
03

Prove W(n) is Cyclic

A group is cyclic if there exists an element \( \zeta \) in \( W(n) \) that can generate the entire group by its powers. The \( n \)-th roots of unity on the complex plane are evenly spaced around the unit circle and are given by \( \zeta = e^{2\pi i / n} \). Consequently, \( \zeta^u \), where \( 0 \leq u < n \), generates all the elements of \( W(n) \), proving it is cyclic with order \( n \).
04

Primitive Roots

A primitive \( n \)-th root of unity is an element \( \zeta \) such that when raised to consecutive powers, generates every element in \( W(n) \). The roots \( \zeta \) for which \( \zeta^d \) is still primitive must satisfy \( \gcd(d,n) = 1 \).
05

Count Primitive Roots

By Euler's Totient Function \( \varphi(n) \), which counts the integers up to \( n \) that are coprime to \( n \), the number of primitive \( n \)-th roots of unity is given by \( \varphi(n) \).

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Key Concepts

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

Unit Roots
In the realm of complex numbers, unit roots, or nth roots of unity, are fascinating mathematical structures. They are complex numbers that, when raised to a certain power, result in 1. Specifically, the set of n-th unit roots is defined as \( W(n) = \{ z \in \mathbb{C} \, ; \, z^n = 1 \} \). These roots form a perfect wheel around the origin of the complex plane, precisely spaced around a unit circle.
Unit roots have interesting properties, primarily because they have an order equal to the integer \( n \). This means there are \( n \) distinct elements in this set, each contributing to the structure of this group. Each nth root of unity is expressed in the exponential form as \( \zeta = e^{2\pi i / n} \). Below are some key properties of unit roots:
  • They include the necessary geometric symmetry on the complex plane.
  • These roots also serve as important elements in fields like Fourier transforms and polynomial factorization.
  • Unit roots always form cyclic groups.
Complex Numbers
Complex numbers can be seen as an extension of the real numbers expressed as pairs, often visualized in the 2D plane, \( \mathbb{R}^2 \). A complex number is represented in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit satisfying \( i^2 = -1 \). This representation opens up a wide array of mathematical operations and solutions that are not possible with just real numbers.
In a geometrical approach, complex numbers correspond to points in the plane, making operations like addition straightforward as vector addition. Multiplication of complex numbers is fascinating due to its rotation and scaling effects on the complex plane, closely related to both angle and magnitude. Key points about complex numbers include:
  • They can be plotted on the Argand diagram, where the x-axis represents the real part, and the y-axis represents the imaginary part.
  • Complex conjugates, found by negating the imaginary part, are critical in reducing polar expressions and finding real results from products.
  • Complex numbers are the building blocks for working with unit roots, Euler's formula, and Fourier analysis.
Euler's Totient Function
Euler’s Totient Function, symbolized as \( \varphi(n) \), helps count the number of positive integers up to \( n \) that are coprime with \( n \). In other words, it measures how many integers less than or equal to \( n \) do not share any factors with \( n \) except for 1. This function is crucial for understanding and calculating primitive roots.
In the context of cyclic groups and unit roots, Euler's Totient Function provides the count of primitive n-th roots of unity. A primitive root is a unit root that cannot be expressed as a power of another root within the set, except for its own powers that generate all the elements in the group. An essential aspect of \( \varphi(n) \) involves its use in various branches of number theory, including:
  • Determining the structure of the multiplicative group of integers modulo \( n \).
  • Utilizing it in the proof of Fermat's Little Theorem.
  • Facilitating calculations related to the RSA encryption algorithm, which is critical in cryptography.
Within each specific application, \( \varphi(n) \) plays a vital role in ensuring comprehensiveness and accuracy.

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Most popular questions from this chapter

Sei \(P\) ein Polynom mit komplexen Koeffizienten: \(P(z):=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{0}\) mit \(n \in \mathbb{N}_{0}, a_{\nu} \in C\), für \(0 \leq \nu \leq n\) Eine reelle oder komplexe Zahl \(\zeta\) heiBt Nultstelle von \(P\), falls \(P(\zeta)=0\) gilt. Man zexge: Wenn alle Koeffirienten \(a_{\nu}\) reell sind, dann gilt $$ P(\zeta)=0 \Longrightarrow P(\bar{\zeta})=0 $$ Mit anderen Worten: Hat das Polynom \(P\) nur reelle Koeffizienten, dann treten die nicht reellen Nullstellen von \(P\) in Paaren konjugiert komplexer Nullstellen auf.

Man verifiziere fir \(z=x+\mathrm{i} y \in \mathrm{C}\) die Ungleichungen und $$ \frac{|x|+|y|}{\sqrt{2}} \leq|z|=\sqrt{x^{2}+y^{2}} \leq|x|+|y| $$ $$ \max \\{|x|,|y|\\} \leq|z| \leq \sqrt{2} \max \\{|x|,|y|\\} $$

Existenz von \(n\)-ten Wurzeln Sei \(a \in C\) und \(n \in N\). Eine komplexe Zahl z heiBt (eine) \(n\)-te Wurzel aus \(a\), wenn \(z^{\mathrm{n}}=a\) gilt. Man zeige: Ist \(a=r(\cos \varphi+i \sin \varphi) \neq 0\), dann besitat a genau \(n\) (verschiedene) \(n\)-te Wurzeln, nämlich die komplexen Zahlen$$ z_{v}=\sqrt[n]{r}\left(\cos \frac{\varphi+2 \pi \nu}{n}+i \sin \frac{\varphi+2 \pi \nu}{n}\right), \quad 0 \leq \nu \leq n-1 $$ Im Spezialfall \(a=1\) (also \(r=1, \varphi=0\) ) erhält man Satz 1.7.

Man bestimme alle \(z \in C\) mit $$ z^{3}-i=0 $$

a) Sei Hl \(:=\\{z \in \mathrm{C} ; \quad \operatorname{lm} z>0\\}\) die obere Halbebene. Man zeige: \(z \in \mathrm{H} \Leftrightarrow-1 / z \in \mathrm{H}\) b) Seien \(z, a \in \mathbb{C}\) Man zeige: \(\quad|1-z \bar{a}|^{2}-|z-a|^{2}=\left(1-|z|^{2}\right)\left(1-|a|^{2}\right) .\) Mon folgere: Ist \(|a|<1\), dann gilt \(|z|<1 \Longleftrightarrow\left|\frac{z-a}{\bar{a} z-1}\right|<1\) und \(|z|=1 \Longleftrightarrow\left|\frac{z-a}{\bar{a} z-1}\right|=1\)

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