# Explaining Complex Integration: If \Gamma Is the Closed Path

• latentcorpse
In summary, we are trying to find a way to prove that the integral of f(z) around the closed path \Gamma is equal to 0. One approach is to find a domain U that is also a semi-annulus in the upper half plane, with an outer radius much greater than R and an inner radius infinitesimally small. This domain is star shaped and we can use Cauchy's Theorem to show that the integral is 0. However, it may not be a rigorous definition of U. Another approach is to use the contour moving technique, where we continuously move the contour without crossing any discontinuities of the function. As long as the function is analytic inside and on the curve, the integral remains
latentcorpse
If $\Gamma$ is the closed path as follows:
from $\delta$ to R along the positive real axis then around the semi circle of radius -R on the upper half plane to -R on the negative real axis then along the negative real axis to $-\delta$ then around the semi circle of radius $\delta$ in the upper half plane and back to $\delta$ on the positive real axis.

If $f(z)=\frac{1-e^{iz}}{z^2}$, explain why $\int_{\Gamma} f(z) dz=0$?

I was thinking of finding a domain,U, for f that was also a semi-annulus in the upper half plane but with outer radius much greater than R (say 1000R) and inner radius infinitesimaly small. Then U is star shaped if we pick the star centre at (0,1000R). $f:U \rightarrow \mathbb{C}$ will be holomorphic and so we can use Cauchy's Theorem to give the answer - but it doesn't seem like a very rigorous definition of U.

Isn't f(z) already holomorphic on the domain in the contour? Why do you want to define another contour?

latentcorpse said:
If $\Gamma$ is the closed path as follows:
from $\delta$ to R along the positive real axis then around the semi circle of radius -R on the upper half plane to -R on the negative real axis then along the negative real axis to $-\delta$ then around the semi circle of radius $\delta$ in the upper half plane and back to $\delta$ on the positive real axis.

If $f(z)=\frac{1-e^{iz}}{z^2}$, explain why $\int_{\Gamma} f(z) dz=0$?

I was thinking of finding a domain,U, for f that was also a semi-annulus in the upper half
plane but with outer radius much greater than R (say 1000R) and inner radius infinitesimaly small. Then U is star shaped if we pick the star centre at (0,1000R). $f:U \rightarrow \mathbb{C}$ will be holomorphic and so we can use Cauchy's Theorem to give the answer - but it doesn't seem like a very rigorous definition of U.

$$f(z)= \frac{1- e^{iz}}{z^2}$$ is analytic everywhere except z= 0 and that is outside the original path. Changing the diameter of the the two circles wont' change that.

Is the problem that the interior is not "star shaped" (there exist at least one point such that the straight line from that point to every other point is in the region)?

First, no matter how large you make one circle or how small you make the other, it will never be "star shaped". Second, Cauchy's Theorem does not apply only to star shaped regions.

the only part of my notes i can find that says cauchy's theorem doesn't need a star shaped domain is called the "contour moving technique"? but that doesn't seem much use here...

anybody?

What's the 'contour moving technique'?

It's impossible to know "where you are" in your course and so we don't know what you have available to use. But it is true that as long as you move a contour "continuously" without crossing any discontinuities of the function, the integral around that contour remains the same. The integral around any "Jordan curve" (basically no self-intersections) is 0 as long as the function is analytic inside and on the curve.

## 1. What is complex integration?

Complex integration is a mathematical concept that involves calculating the integral of a function over a complex plane, rather than just a real number line. It is used in various fields of science and engineering, such as physics and electrical engineering.

## 2. What is a closed path in complex integration?

A closed path, also known as a closed contour or closed curve, is a continuous curve in the complex plane that starts and ends at the same point. It is used as the boundary for calculating a complex integral.

## 3. What is the role of the path in complex integration?

The path is an essential component in complex integration as it determines the region over which the integral is being calculated. It also affects the value of the integral, as different paths can result in different values for the same function.

## 4. How is complex integration different from real integration?

Complex integration differs from real integration in that it involves integrating over a two-dimensional plane rather than just a one-dimensional line. It also takes into account the direction of integration, whereas in real integration, the direction does not matter.

## 5. What is the importance of complex integration in science?

Complex integration is important in science as it allows us to solve complex mathematical problems that cannot be solved using real integration. It also has applications in various fields, such as signal processing, fluid dynamics, and quantum mechanics.

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