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a) Man betrachte die Abbildung $$ f: \mathrm{C}^{*} \rightarrow \mathrm{C} \text { mit } f(z)=1 / \bar{z} $$ Man gebe eine geometrische Konstruktion (Zirkel und Lineal) für den Bildpunkt \(f(z)\) und begrinde, warum diese Abbildung „Transformation durch reziproke Radien" oder „Spiegelung an der Binheitskreislinie" genannt wird. Man bestimme jeweils das Bild unter \(f\) von \alpha) \(D_{1}:=\\{z \in \mathrm{C} ; \quad 0<|z|<1\\}\), (3) \(D_{2}:=\\{z \in \mathbf{C} ; \quad|z|>1\\}\) \gamma) \(D_{3}:=\\{z \in \mathrm{C} ; \quad|z|=1\\}\) b) Jetat betrachte man die Abbildung $$ g: \mathbb{C}^{*} \rightarrow \mathrm{C} \text { mit } g(z)=1 / z(=\overline{f(z)}) $$ und gebe ebenfalls eine geometrische Konstruktion far den Bildpunkt \(g(z)\) von 2. Warum heißt diese Abbildung „Inversion an der Einheitskreislinie"?

Short Answer

Expert verified
1. f(z) inverts over unit circle: maps |z|<1 to |w|>1, |z|>1 to |w|<1. 2. g(z) conjugate to f(z), maintains symmetry.

Step by step solution

01

Understanding the Mapping

To understand the function \(f(z) = \frac{1}{\bar{z}}\), we need to recognize that it transforms a complex number by taking the reciprocal of its conjugate. In polar form \(z = re^{i\theta}\), its conjugate is \(\bar{z} = re^{-i\theta}\), and hence \(f(z) = \frac{1}{r} e^{i\theta}\). This implies the function maps \(z\) to a point on a circle with a radius of \(\frac{1}{r}\).
02

Geometric Construction of f(z)

For any given \(z\), draw the line from \(z\) to the origin. Measure the distance \(|z|\) and construct a circle centered at the origin with the radius \(\frac{1}{|z|}\). The point where this circle intersects the line extending through \(z\) is the image \(f(z)\). This operation inverts \(z\) across the unit circle and is why it's called 'transformation by reciprocal radii' or 'reflection in the unit circle'.
03

Image of Set D1 under f

For any \(z\) in \(D_1 = \{z \in \mathrm{C} : 0 < |z| < 1\}\), since \(|z| < 1\), \(f(z)\) transforms \(z\) onto a point outside the unit circle: \(|f(z)| = \frac{1}{|z|} > 1\). Thus, \(f(D_1) = \{w \in \mathrm{C} : |w| > 1\}\).
04

Image of Set D2 under f

For any \(z\) in \(D_2 = \{z \in \mathrm{C} : |z| > 1\}\), \(f(z)\) maps \(z\) inside the unit circle because \(|f(z)| = \frac{1}{|z|} < 1\). Therefore, \(f(D_2) = \{w \in \mathrm{C} : 0 < |w| < 1\}\).
05

Image of Set D3 under f

For any \(z\) in \(D_3 = \{z \in \mathrm{C} : |z| = 1\}\), \(f(z)\) maps \(z\) back to itself on the unit circle since \(f(z) = \frac{1}{z}\), and \(|z|=1\) implies \(\left|\frac{1}{z}\right|=1\). Thus, \(f(D_3) = \{w \in \mathrm{C} : |w| = 1\}\).
06

Understanding the Mapping for g

Now, consider the function \(g(z) = \frac{1}{z}\). This is effectively the function \(f(z)\) with an added conjugation step, reversing the angle of \(z\). It can be constructed geometrically by marking the position \(\frac{1}{|z|}\) on the line from the origin through \(z\). As the reflection symmetry is preserved, this function remains an 'inversion in the unit circle' but now with reflected angles.

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Key Concepts

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

Complex Functions
Complex functions are mathematical functions that extend the idea of algebraic functions, like polynomials or exponentials, to the complex plane. These functions map complex numbers to other complex numbers, producing rich geometric visuals in the process.

In the case of functions like \( f(z) = \frac{1}{\bar{z}} \), we transform a number \( z \) through its complex conjugate. The function is well-defined except for zero, meaning it operates on all non-zero complex numbers, denoted as \( \mathbb{C}^* \). By considering transformations in terms of their polar forms, such as \( z = re^{i\theta} \), where \( r \) is the magnitude and \( \theta \) is the angle, complex functions offer a powerful way to manipulate complex numbers under various operations.

Complex analysis uses these functions to explore intricate patterns and mappings, which can be visually represented or physically interpreted, enhancing our grasp of mathematical concepts.
Geometric Inversion
Geometric inversion is a fascinating concept where points in a plane are mapped to other points in such a way that products of distances with respect to a fixed circle (typically a unit circle) are constant. It is a special kind of geometric transformation that can be particularly intuitive and useful.

Consider a point \( z \) outside a unit circle. The geometric inversion will place its image within the circle. The concept revolves around the distance reciprocation from the center of the circle. In our case with \( f(z) = \frac{1}{\bar{z}} \), this inversion maps the distance \( r \) to \( \frac{1}{r} \). Such transformations are deeply studied in complex analysis and can be used to solve complex geometrical problems by reflecting the points across the circle, often turning circles and lines into other circles or lines.
Unit Circle Transformation
Unit circle transformation specifically focuses on mapping points in relation to the unit circle. The unit circle is a fundamental concept in complex analysis and geometry, defined as \( |z| = 1 \). It is the collection of all points in the complex plane that are exactly a unit distance from the origin.

For a function like \( f(z) = \frac{1}{\bar{z}} \), any point \( z \) exactly on the unit circle remains unchanged, illustrating a self-mapping property. This signifies the nature of its critical boundary in transformations. Understanding transformations around the unit circle aids in analyzing behavior like stability and boundary changes in complex systems.

For instance, points inside the unit circle map outside and vice versa, creating dynamic inversions that are symmetric about the circle's circumference. This can be instrumental in fields such as control theory and electrical engineering, where system stability is paramount.
Reciprocal Mapping
Reciprocal mapping involves transforming each point \( z = re^{i\theta} \) into its reciprocal magnitude, but maintaining the direction, or angle, unaffected. The basic idea is reversing distance while retaining orientation, creating a simple yet effective transformation.

More precisely, for the mapping transformation \( f(z) = \frac{1}{\bar{z}} \), this results in flipping the distance from the origin but preserving the angle, hence transforming radius \( r \) to \( \frac{1}{r} \).

Through reciprocal mapping, any complex number can be repositioned, which is incredibly beneficial for certain symmetrical and analytical conditions in complex analysis, offering logical solutions to problems of distance and orientation. Also known as reflecting transformation, reciprocal mappings provide insights into deeper visual interpretations in mathematical modelling.
Polar Coordinates
Polar coordinates are a unique way of representing points in a plane, differing from the usual Cartesian coordinates. They describe a point in terms of its distance from a reference point (the pole, typically the origin) and an angular direction (measured from a fixed line, often the positive x-axis).

In complex numbers, representing a number as \( z = re^{i\theta} \), where \( r \) is the modulus and \( \theta \) the argument, aligns well with polar coordinates. This form simplistically displays a number’s geometry which is especially useful in transformations, like with \( f(z) = \frac{1}{\bar{z}} \).

Using polar coordinates allows for smooth handling of multiplications and divisions, because the modulus is simply scaled, and the angles are added or subtracted directly. This interpretation aids students in visualizing how number magnitudes and angles shift through various transformations, offering a powerful method of understanding complex functions and their relations geometrically.

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Most popular questions from this chapter

Von den folgenden komplexen Zahlen bestimme man jeweils Real- und Imaginärteit: $$ \begin{gathered} \frac{i-1}{i+1} ; \quad \frac{3+4 i}{1-2 i} ; i^{n}, n \in Z ; \quad\left(\frac{1+i}{\sqrt{2}}\right)^{n}, n \in Z \\ \left(\frac{1+i \sqrt{3}}{2}\right)^{n}, n \in Z: \sum_{v=0}^{T}\left(\frac{1-i}{\sqrt{2}}\right)^{v} ; \frac{(1+i)^{4}}{(1-i)^{3}}+\frac{(1-i)^{4}}{(1+i)^{3}} \end{gathered} $$

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

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

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.

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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