# Prove No Analytic Function F on Annulus D: 1<|z|<2

• oab729
In summary, an analytic function is a complex-valued function that is differentiable at every point in its domain. An annulus is a region in the complex plane bounded by two concentric circles, and it is important to prove that there is no analytic function on this annulus to understand the behavior of analytic functions in different regions. This can be proven by contradiction, and the implications of this result include the existence of regions in the complex plane where analytic functions do not exist and the impact of singularities on the behavior of these functions. This result is also useful in the study of complex analysis and its applications in various fields.
oab729

## Homework Statement

Prove that there does not exist an analytic function on the annulus D: 1<|z|<2, s.t. F'(z) = 1/z for all z in D.

## The Attempt at a Solution

Assume F exists, then for z in D, not a negative number, F(z) = Log z + c since Log' z = 1/z... Lost

Find a path C such that $$\int_C \frac1z\,dz\neq0$$

## 1. What is an analytic function?

An analytic function is a complex-valued function that is differentiable at every point in its domain. This means that the function has a well-defined slope at each point, and the value of the function can be approximated by a polynomial.

## 2. What is an annulus?

An annulus is a region in the complex plane that is bounded by two concentric circles. In this case, the annulus D is defined by the inequality 1 < |z| < 2, which means that it includes all points with a distance of at least 1 from the origin, but less than 2.

## 3. Why is it important to prove that there is no analytic function on this annulus?

Proving that no analytic function exists on this annulus is important because it helps us understand the behavior of analytic functions in different regions of the complex plane. It also has practical applications, such as in the study of potential theory and conformal mapping.

## 4. How can we prove that no analytic function exists on this annulus?

We can prove this by contradiction. Assume that there exists an analytic function f on the annulus D. Then, using Cauchy's integral theorem, we can show that the integral of f along a closed contour in the annulus is equal to 0. However, by choosing a specific contour, such as a circle with radius 1.5, we can show that this integral is not equal to 0, leading to a contradiction.

## 5. What are the implications of this result?

This result implies that there are certain regions in the complex plane where analytic functions do not exist. It also shows that analytic functions can have very different properties in different regions, and that the presence or absence of singularities can greatly impact the behavior of these functions. This result is also useful in the study of complex analysis and its applications in physics, engineering, and other fields.

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