Complex analysis antiderivative existence

In summary, f(z)=1/z and f(z)=(1/z)^n do not have antiderivatives over the domain C/(0,0) due to the non-zero integral over a closed contour, making them not path independent.
  • #1
reb659
64
0

Homework Statement



a) Does f(z)=1/z have an antiderivative over C/(0,0)?

b) Does f(z)=(1/z)^n have an antiderivative over C/(0,0), n integer and not equal to 1.

Homework Equations


The Attempt at a Solution



a) No. Integrating over C= the unit circle gives us 2*pi*i. So for at least one closed contour the integral is nonzero. Thus f(z) cannot be path independent and thus cannot have an antiderivative over the domain.

b) By using the same reasoning it seems that no antiderivative over the domain exists either, but I'm not sure.
 
Last edited:
Physics news on Phys.org
  • #2
reb659 said:

Homework Statement



a) Does f(z)=1/z have an antiderivative over C/(0,0)?

b) Does f(z)=(1/z)^n have an antiderivative over C/(0,0), n integer and not equal to 1.

Homework Equations





The Attempt at a Solution



a) No. Integrating over C= the unit circle gives us 2*pi*i. So for at least one closed contour the integral is nonzero. Thus f(z) cannot be path independent and thus cannot have an antiderivative over the domain.

b) By using the same reasoning it seems that no antiderivative over the domain exists either, but I'm not sure.

For b) why don't you make sure by integrating over a contour? Try z=e^(i*t) for t in [0,2pi].
 

FAQ: Complex analysis antiderivative existence

1. What is complex analysis antiderivative existence?

Complex analysis antiderivative existence is a concept in mathematics that refers to the existence of antiderivatives, also known as primitive functions, for complex-valued functions. It is a fundamental concept in complex analysis and plays a crucial role in the study of complex functions.

2. Why is the existence of antiderivatives important in complex analysis?

The existence of antiderivatives is important in complex analysis because it allows us to find the indefinite integral of complex functions. This, in turn, helps us to evaluate complex integrals, which are essential in many areas of mathematics and physics, such as in the calculation of areas, volumes, and physical quantities.

3. How can we determine if a complex function has an antiderivative?

A complex function has an antiderivative if it satisfies the Cauchy-Riemann equations, which are necessary and sufficient conditions for the existence of an antiderivative. These equations are based on the differentiability of the function and its continuity in a given region of the complex plane.

4. Can all complex functions have an antiderivative?

No, not all complex functions have an antiderivative. For a complex function to have an antiderivative, it must satisfy the Cauchy-Riemann equations and be continuous in a given region of the complex plane. If a function fails to meet these conditions, it will not have an antiderivative.

5. How is the existence of antiderivatives related to the concept of analyticity?

The existence of antiderivatives is closely related to the concept of analyticity in complex analysis. A complex function is said to be analytic in a region if it has an antiderivative in that region. Therefore, the existence of antiderivatives is a key criterion for a function to be analytic, which has many important consequences in complex analysis.

Similar threads

Can I Use Antiderivatives to Evaluate this Complex Integral?
Replies
7
Views
2K
Contour integration around a complex pole
Replies
2
Views
2K
[Complex analysis] Contradiction in the definition of a branch
Replies
8
Views
2K
Complex Analysis: Contour Integration Question
Replies
1
Views
2K
Complex Integration Along Given Path
Replies
3
Views
1K
Is f(z) =(1+z)/(1-z) a real function?
Replies
17
Views
2K
Show that the real part of a certain complex function is harmonic
Replies
7
Views
2K
Is this question incomplete? Regarding entire functions....
Replies
1
Views
1K
Proof given ##x < y < z## and a twice differentiable function
Replies
8
Views
917
Proof showing that if F is an antiderivative of f, then f must be continuous.
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top