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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.

Short Answer

Expert verified
Non-real roots of a real-coefficient polynomial occur in conjugate pairs.

Step by step solution

01

Understand the Problem Statement

We are given a polynomial \(P(z)\) with complex coefficients. A number \(\zeta\) is a root of the polynomial if \(P(\zeta) = 0\). We need to prove that if all coefficients \(a_u\) are real, then for any root \(\zeta\), its complex conjugate \(\bar{\zeta}\) is also a root.
02

Write Down the Polynomial Equations

Consider the polynomial \(P(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_0\), where all coefficients \(a_n, a_{n-1}, \ldots, a_0\) are real.
03

Apply the Complex Conjugate

Take the complex conjugate of both sides of the polynomial equation \(P(\zeta) = 0\). Thus, we have:\[\overline{P(\zeta)} = \bar{0} = 0.\]This means \(\overline{a_n \zeta^n + a_{n-1} \zeta^{n-1} + \cdots + a_0} = 0\).
04

Use Properties of Complex Conjugates

The complex conjugate of a sum is the sum of the complex conjugates, and \((ab)^* = a^* b^*\), where \(a^*\) is the conjugate of \(a\). So, \[\overline{P(\zeta)} = \overline{a_n \zeta^n} + \overline{a_{n-1} \zeta^{n-1}} + \cdots + \overline{a_0} = \overline{a_n} \cdot \overline{\zeta^n} + \cdots + \overline{a_0}.\]Since all \(a_u\) are real, \(\overline{a_u} = a_u\).
05

Simplify Using Real Coefficients

Since \(a_u\) are real, we simplify: \[P(\bar{\zeta}) = a_n \overline{\zeta}^n + a_{n-1} \overline{\zeta}^{n-1} + \cdots + a_0 = a_n (\bar{\zeta})^n + a_{n-1} (\bar{\zeta})^{n-1} + \cdots + a_0.\]Comparing it to the expression \(\overline{P(\zeta)}\) from step 4, we have \[P(\bar{\zeta}) = \overline{P(\zeta)} = 0.\]So, \(P(\bar{\zeta}) = 0\).
06

Conclusion: Conjugate Root Theorem

Thus, we have shown \(P(\zeta) = 0\) implies \(P(\bar{\zeta}) = 0\) when all coefficients are real, confirming that non-real roots occur in conjugate pairs.

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Key Concepts

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

Polynomials with Real Coefficients
Polynomials are expressions consisting of variables and coefficients, combined using addition, subtraction, and multiplication. When these coefficients are real numbers, we refer to such polynomials as having "Polynomials with Real Coefficients". These polynomials are fundamental in advanced mathematics and can be understood as natural extensions of linear functions.

Real coefficients in polynomials have intriguing implications, especially when determining the nature of the roots. If a polynomial has only real coefficients, then any non-real root must occur in conjugate pairs. This curious property is an integral part of understanding polynomial behavior and helps when solving polynomial equations using the Conjugate Root Theorem.

A polynomial of degree \(n\) generally has \(n\) roots, which can include real numbers and pairs of complex conjugates. This relationship arises because complex numbers have both a real part and an imaginary part, and their behavior is symmetrical around the real axis of the complex plane, which directly ties to their appearance in conjugate pairs.
Complex Conjugates
Complex numbers are numbers that include both a real part and an imaginary part, typically in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with \(i^2 = -1\).

A complex conjugate is formed by changing the sign of the imaginary part of a complex number. For example, the complex conjugate of \(a + bi\) is \(a - bi\).

Some essential properties of complex conjugates are:
  • The complex conjugate of a sum or product of complex numbers is the sum or product of their conjugates.
  • The product of a complex number and its conjugate is always a real number. Specifically, \((a+bi)(a-bi) = a^2 + b^2\), which is a positive real number.
  • Complex conjugates have the same magnitude; hence they lie symmetrically about the real axis.
Understanding these properties is vital, particularly when examining polynomials with real coefficients, because it enables us to predict the presence of conjugate pairs among the roots.
Roots of Polynomials
In mathematical terms, a root (or zero) of a polynomial is a value for which the polynomial equals zero. If \(\zeta\) is a root of a polynomial \(P(z)\), then \(P(\zeta) = 0\).

Roots can be real or complex, and their nature depends on the coefficients of the polynomial and the polynomial's degree. For polynomials with real coefficients, a fascinating characteristic is that any non-real roots must come in pairs of complex conjugates. This is where the Conjugate Root Theorem comes into play.

The theorem asserts that if \(P(z)\) is a polynomial with real coefficients, then any non-real complex root \(\zeta\) also implies that its complex conjugate \(\bar{\zeta}\) is a root. This results in an even count of non-real roots in a polynomial. Understanding this helps simplify solving polynomial equations and predicting the roots' behavior.

Moreover, the total number of roots (considering multiplicity) of a polynomial is equal to its highest degree. This theorem aids not only in understanding the nature of roots but also in visualizing them on the complex plane, enhancing our overall comprehension of polynomial equations.

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Most popular questions from this chapter

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 \(|z|=1 \Longleftrightarrow\left|\frac{z-a}{\bar{a} z-1}\right|=1\)

Quadratwurzeln und Lösbarkeit quadratischer Gleichungen in C Sei \(c=a+i b \neq 0\) eine vorgegebene komplexe Zahl. Durch Aufspaltung in Realund Imaginärteil zeige man, dass es genau zwei verschiedene komplexe Zahlen \({ }_{1} \mathrm{~s}\) und \(z_{2}\) gibt mit $$ z_{1}^{2}=t_{2}^{2}=c . \text { Es ist } z_{2}=-z_{1^{*}} $$ \(\left(z_{1}\right.\) und \(z_{2}\) heiben die Qundratwurzeln aus c.) Als Beispiel bestimme man jeweils die Quadratwurzeln aus. $$ 5+7 i \text { bzw. } \sqrt{2}+i \sqrt{2} $$ Man löse diese Aufgabe auch mit Polarkoordinaten. Ferner zeige man, dass eine quadratische Gleichung $$ x^{2}+\alpha z+\beta=0, \quad \alpha, \beta \in \mathbb{C} \text { beliebig, } $$ stets (h?chstens zwei) Lösungen \(z_{1}, z_{2} \in \mathbb{C}\) besitzt.

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 beweise die \(„\) Dreiecksungleich wng" $$ |z+w| \leq|z|+|\mathbf{w}|, \quad z, w \in C $$ und diskutiere, wann das Gleichheitszeichen gilt; ferner beweise man die folgende Variante der Dreiecksungleichung: $$ || z|-| w|| \leq|z-w|, \quad 2, w \in \mathbb{C} $$

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

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