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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: $$ || z|-| w|| \leq|z-w|, \quad 2, w \in \mathbb{C} $$

Short Answer

Expert verified
The triangle inequality \(|z+w| \leq |z| + |w|\) holds for any complex numbers and equality occurs when \(z\) and \(w\) are collinear. The variant \(||z| - |w|| \leq |z-w|\) holds by similar reasoning.

Step by step solution

01

Understand the Triangle Inequality

The triangle inequality states that for any complex numbers \(z\) and \(w\), the magnitude of their sum is less than or equal to the sum of their magnitudes. Mathematically, it is expressed as \(|z+w| \leq |z| + |w|\). This is a principal property of normed vector spaces.
02

Prove the Triangle Inequality

To prove the inequality \(|z+w| \leq |z| + |w|\), consider \((z + w)\overline{(z + w)}\) which simplifies to \(|z|^2 + |w|^2 + 2\text{Re}(z\overline{w})\). Then using the Cauchy-Schwarz inequality for complex numbers, \(2\text{Re}(z\overline{w}) \leq 2|z||w|\), we find \(|z|^2 + |w|^2 + 2\text{Re}(z\overline{w}) \leq (|z| + |w|)^2\). Thus, \(|z+w| \leq |z| + |w|\).
03

Discuss Equality Condition

The equality \(|z+w| = |z| + |w|\) holds if and only if \(z\) and \(w\) have the same direction, i.e., they are positive scalar multiples of each other. In other words, \(z\) and \(w\) lie on the same straight line originating from the origin in the complex plane.
04

Prove the Variant of Triangle Inequality

To prove the second inequality \(||z| - |w|| \leq |z-w|\), use the previously proven inequality and consider: \(|z| = |z-w + w| \leq |z-w| + |w|\). Similarly, \(|w| = |w-z + z| \leq |w-z| + |z| = |z-w| + |z|\). Thus, \(|z| - |w| \leq |z-w|\) and \(|w| - |z| \leq |z-w|\) imply \(||z| - |w|| \leq |z-w|\).
05

Discuss Variant Equality Condition

The equality \(||z| - |w|| = |z-w|\) holds when \(z\) and \(w\) are equidistant from the origin. This happens when \(z\) and \(w\) lie on a straight line that passes through the origin and \(z-w\) is the vector between two points on the line.

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Key Concepts

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

Triangle Inequality
The Triangle Inequality is a fundamental principle in complex analysis and mathematics in general. This inequality tells us that for any two complex numbers, say \(z\) and \(w\), the absolute value (or magnitude) of their sum is always less than or equal to the sum of their magnitudes.
  • Formally, it is written as \(|z + w| \leq |z| + |w|\).
  • This property can help us understand the 'shape' or 'geometry' within normed vector spaces.

In simple terms, imagine two points in the complex plane. The straight "line" that connects them is always shorter or equal to the path that moves between them in two segments. When does equality hold? If and only if \(z\) and \(w\) point in the exact same direction or one of them is just a stretched or compressed version of the other.
Through equations, if \(z = kw\) where \(k\) is a positive number, then their sum equals their individual magnitudes. This geometric intuition helps solve many problems related to distances and magnitudes in complex spaces.
Complex Numbers
Complex numbers are at the core of understanding many mathematical and engineering concepts. A complex number is formed of two parts:
  • Real part – denoted usually by \(a\)
  • Imaginary part – denoted by \(bi\), where \(i\) is the imaginary unit with the property \(i^2=-1\)
Together, a complex number looks like \(a + bi\).
The beauty of complex numbers lies in their ability to represent both magnitude and direction.
When represented graphically in a complex plane:
  • The real part determines horizontal placement.
  • The imaginary part determines vertical placement.

Furthermore, their magnitude, crucial for inequalities, is determined by the formula \(\sqrt{a^2 + b^2}\). Understanding complex numbers allows us to extend real number concepts into two dimensions, providing a richer framework for solving equations that no real solution could answer.
Normed Vector Spaces
In mathematics, a normed vector space is a vector space with a function, called a norm, which assigns a strictly positive length or size to each vector in the space (other than the zero vector). The idea is essential in areas of analysis.
  • The norm, typically denoted \(||v||\), is akin to the 'length' of a vector \(v\) in the space.
  • Examples include the Euclidean norm in context of spatial vectors or the magnitude of complex numbers.

These spaces are essential for defining and understanding the Triangle Inequality in more abstract terms.
In normed vector spaces, the triangle inequality's importance is vast. It essentially says that the "straight line" path in a vector space doesn't exceed the magnitude-sum of alternate paths.By mastering the concept of normed spaces, we lay the foundation for more complex analyses, understanding convergence and continuity of vectorial functions in infinite dimensions.
Cauchy-Schwarz Inequality
The Cauchy-Schwarz Inequality is a powerful mathematical tool widely used in linear algebra and complex analysis. It provides a way of measuring the 'angle' between two vectors and is expressed for complex numbers as:
  • \(| \langle z, w \rangle | \leq ||z|| ||w||\)

This inequality underpins the reasoning used in proving the Triangle Inequality. The idea behind it is comparing products of vectors and magnitudes, illustrating how vectors project onto each other.
  • It's a statement about dot products (or inner products) of vectors.
  • The equality in Cauchy-Schwarz holds when \(z\) and \(w\) are linearly dependent – they essentially lie along the same line.

Understanding and applying the Cauchy-Schwarz Inequality helps in grasping relations in vector spaces more deeply and gaining insights into the geometrical properties of solutions.

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Most popular questions from this chapter

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.

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.

Jedes \(z \in S^{1}-\\{-1\\}\), \(S^{1} ;=\\{z \in \mathrm{C} ; \quad|z|=1\\}\) l?sst sich eindeutig in der Form \(z=\frac{1+i \lambda}{1-i \lambda}=\frac{1-\lambda^{2}}{1+\lambda^{2}}+\frac{2 \lambda}{1+\lambda^{2}} i\) mit \(\lambda \in \mathbb{R}\) darstellen.

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

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.

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