Problem 1 - Arbeitsbuch Mathematik für Ingenieure

Question

Sei \(f(z)=\frac{1}{2}\left(z+\frac{1}{z}\right), z \in \mathbb{C} \backslash\\{0\\}\). a) Bestimmen Sie Real- und Imaginärteil dieser Funktion. b) Bestimmen Sie die Bildmenge der imaginären Halbachse \(\operatorname{Re} z=0, \operatorname{Im} z>0\) c) Bestimmen Sie die Bildmenge des Einheitskreises \(|z|=1\). d) Bestimmen Sie alle Nullstellen der Funktion.

Step-by-Step Solution

Answer

a) Real: \frac{1}{2}(\cos\theta)(r + r^{-1}), Imaginary: \frac{1}{2}(\sin\theta)(r - r^{-1})\ b) Image: i\mathbb{R}\ c) Image: [-1, 1]\ d) Zeros: \pm i\.

Step 1: Find the Real and Imaginary Parts of the Function

Given the function \(f(z) = \frac{1}{2}(z + \frac{1}{z})\), write \(z\) in its polar form: \(z = re^{i\theta}\) where \(r\) is the modulus and \(\theta\) is the argument.\( \frac{1}{z} = \frac{1}{r}e^{-i\theta}\). Hence, \(f(z) = \frac{1}{2} (re^{i\theta} + r^{-1}e^{-i\theta})\). Now, separate into real and imaginary parts using Euler's formulas: \(re^{i\theta} = r(\cos \theta + i \sin \theta)\) and \(r^{-1}e^{-i\theta} = r^{-1}(\cos \theta - i \sin \theta)\). Combine these to get real: \((\cos\theta)(r + r^{-1})\) and imaginary:\((\sin\theta)(r - r^{-1})\) parts.

Step 2: Determine the Image of the Imaginary Axis

For \(z = iy\) where \(y > 0\), substitute \(iy\) into the function:\( f(iy) = \frac{1}{2} (iy + \frac{1}{iy}) = \frac{1}{2} (iy - i \frac{1}{y}) = \frac{1}{2} i (y - \frac{1}{y}) \). Hence, the image of the imaginary axis is: \ f(iy) \in i \mathbb{R}\.

Step 3: Determine the Image of the Unit Circle

For \(|z| = 1\), let \(z = e^{i\theta}\). Substitute into the function: \( f(e^{i\theta}) = \frac{1}{2} (e^{i\theta} + e^{-i\theta}) = \frac{1}{2} (2\cos\theta) = \cos\theta \). Hence, the image of the unit circle is the interval \[-1, 1\].

Step 4: Find the Zeros of the Function

To find where \(f(z) = 0\): \(f(z) = \frac{1}{2}(z + \frac{1}{z}) = 0\). This simplifies to \ z + \frac{1}{z} = 0\ or \ z^{2} = -1 \. Thus, the solutions are \(z = \pm i\).

Key Concepts

Real and Imaginary Parts

When we talk about complex functions, identifying the real and imaginary parts is crucial. In this exercise, we start with the function given by:
\( f(z) = \frac{1}{2} (z + \frac{1}{z}) \).
We express the complex number \(z\) in its polar form:
\( z = re^{i\theta} \), where \(r\) is the modulus (distance from the origin in the complex plane), and \( \theta \) is the argument (angle with the real axis).
This means:
\( \frac{1}{z} = \frac{1}{r}e^{-i\theta} \).
We substitute back into the function, yielding
\( f(z) = \frac{1}{2} (re^{i\theta} + r^{-1}e^{-i\theta}) \).
Using Euler's formulas:
Euler's formula states:
\( e^{i\theta} = \cos \theta + i \sin \theta \) and \( e^{-i\theta} = \cos \theta - i \sin \theta \).
Substituting these in, we get:
\( f(z) = \frac{1}{2} \left[ r(\cos \theta + i \sin \theta) + r^{-1}(\cos \theta - i \sin \theta) \right] \).
Collecting real and imaginary parts:
Real part: \( \text{Re} f(z) = \frac{1}{2} [(r + r^{-1}) \cos \theta] \),
Imaginary part: \( \text{Im} f(z) = \frac{1}{2} [(r - r^{-1}) \sin \theta] \).

Image of a Complex Function

To understand where the points from one set map to another under a complex function, we investigate the image of a complex function.
For part b, we focus on the imaginary axis where \( \text{Re} \,z = 0 \) and \( \text{Im} \,z > 0 \). Taking \( z = iy \) (where y is a real positive number), substituting \(z = iy \) into the function \( f(z) \), gives:
\( f(iy) = \frac{1}{2} (iy + \frac{1}{i y}) = \frac{1}{2} (iy - \frac{i}{y}) \).
Simplifying, we get:
\( f(iy) = \frac{1}{2} i ( y - \frac{1}{y}) \).
This result is purely imaginary since it takes the form \( i c \) where \( c \) is a real number, meaning we have:
\( f(iy) \in i \mathbb{R} \).
Thus, the image of the imaginary axis is purely an imaginary set of points described by:
\( f(iy) \in i \mathbb{R} \).

Unit Circle Mapping

Mapping the unit circle using complex functions is essential in understanding their periodic behavior.
Here we investigate the image of the function when \( |z| = 1 \). For a complex number \( z \) on the unit circle, we have:
\( z = e^{i \theta} \), where \( \theta \) ranges from 0 to \( 2 \pi \).
Substituting \( z = e^{i \theta} \) into the function:
\( f(e^{i \theta}) = \frac{1}{2} (e^{i \theta} + e^{-i \theta}) \).
Using Euler’s formulas again:
\( e^{i \theta} + e^{-i \theta} = 2 \cos \theta \).
So the function simplifies to:
\( f(e^{i \theta}) = \cos \theta \).
Since \( \cos \theta \) ranges between -1 and 1, the image of the unit circle under the function \( f \) is the interval:
\[ -1, 1 \].

Zeros of Functions

Finding the zeros of a complex function helps in understanding its critical points. We need to determine where \( f(z) = 0 \). For our given function:
\( f(z) = \frac{1}{2} (z + \frac{1}{z}) \), setting it to zero, we solve:
\( \frac{1}{2} (z + \frac{1}{z}) = 0 \).
Simplifying, this becomes:
\( z + \frac{1}{z} = 0 \).
Solving this equation, we multiply through by \( z \), giving: \( z^{2} = -1 \).
Thus, the solutions are the imaginary units:
\( z = \pm i \).
Hence, the zeros of the function \( f(z) \) are at the points:
\( +i \) and \( -i \).