Chapter 10: Problem 6
Weisen Sie nach, dass $$ F(z)=\log z=\ln \sqrt{x^{2}+y^{2}}+i \arctan \frac{y}{x}, x \neq 0 $$ eine Stammfunktion von \(f(z)=\frac{1}{z}\) ist.
Short Answer
Expert verified
\( F(z) = \log z \) is an antiderivative of \( f(z) = \frac{1}{z} \).
Step by step solution
01
Understand the Definition of a Antiderivative
An antiderivative or primitive function of a function \( f(z) \) is a function \( F(z) \) such that \( F'(z) = f(z) \). We need to prove \( F(z) = \log z = \ln \sqrt{x^2+y^2} + i \arctan \frac{y}{x} \) satisfies this condition for \( f(z) = \frac{1}{z} \).
02
Express \( z \) as a Complex Number
A complex number \( z \) can be expressed in Cartesian form as \( z = x + iy \). Therefore, the modulus is \( |z| = \sqrt{x^2 + y^2} \) and the argument is \( \text{arg}(z) = \arctan \frac{y}{x} \), given \( x eq 0 \).
03
Differentiate \( F(z) = \log z \)
Differentiate \( F(z) \) with respect to \( z \). Since \( \log z \) in complex analysis is defined as \( \log z = \ln |z| + i \text{arg}(z) \), we have: \(\frac{d}{dz} \log z = \frac{1}{z}\)
04
Verify the Derivative Matches \( f(z) \)
We calculated \( F'(z) = \frac{1}{z} \). This matches the function \( f(z) \) given in the problem. Therefore, \( F(z) \) is indeed an antiderivative of \( f(z) = \frac{1}{z} \).
05
Conclude that \( F(z) \) is an Antiderivative
Since the derivative \( F'(z) \) equals \( f(z) = \frac{1}{z} \), we have shown that \( F(z) = \log z \) is a valid antiderivative. This completes the proof.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
An antiderivative, also known as a primitive function, is a fundamental concept in calculus, including complex analysis. Simply put, an antiderivative of a function \( f(z) \) is another function \( F(z) \) such that when you differentiate \( F(z) \), you get \( f(z) \). This gives us the relation:
- \( F'(z) = f(z) \)
Logarithm
The logarithm in complex analysis extends the idea of logarithms used in real analysis. For a complex number \( z \), the logarithm \( \log z \) can be expressed as:
- \( \log z = \ln |z| + i \text{arg}(z) \)
Complex Numbers
Complex numbers, denoted as \( z = x + iy \), consist of a real part \( x \) and an imaginary part \( y \), where \( i \) is the imaginary unit with the property that \( i^2 = -1 \).
- Cartesian form of a complex number: \( z = x + iy \)
- Polar form: \( z = |z|(\cos \theta + i\sin \theta) \)
Differentiation
Differentiation in complex analysis is a natural extension of the differentiation we learn with real numbers. Here, when we differentiate a complex function \( F(z) \), we ensure that the derivative matches the given function in question. In our exercise, differentiating \( F(z) = \log z \) involves recognizing how the complex logarithm behaves:
- \( \frac{d}{dz} \log z = \frac{1}{z} \)
Modulus and Argument
The modulus and argument are crucial for understanding and working with complex numbers. The modulus \( |z| \) of a complex number \( z = x + iy \) is the distance from the origin to the point \( (x, y) \) in the complex plane, calculated as:
- \( |z| = \sqrt{x^2 + y^2} \)
- \( \text{arg}(z) = \arctan \frac{y}{x} \)