# Contour Integration over Square, Complex Anaylsis

• Safder Aree
In summary, the path from 0 to 1 is z(t) = t, z'(t) = 1, and the path from 1 to 1+i is z(t) = 1+i-t, z'(t) = -1.
Safder Aree

## Homework Statement

Show that
$$\int_C e^zdz = 0$$

Let C be the perimeter of the square with vertices at the points z = 0, z = 1, z = 1 +i and z = i.

## Homework Equations

$$z = x + iy$$

## The Attempt at a Solution

I know that if a function is analytic/holomorphic on a domain and the contour lies within the domain lies. Then this integral equals 0. I can prove that e^z is analytic everywhere. (using cauchy-riemann) Is that enough to show it works in this case?

Let me make the statement more correct. Cauchy's theorem says that an analytic function has integral zero over a closed contour, lying in its region of analyticity, provided that contour does not wind around any point where the function is not analytic. In the case of a simple closed contour like this one, that means the function must be analytic on the contour as well as on the points interior to the contour. Does that apply?

In this extremely simple case, it might be more instructive to actually parametrize the contour and compute the integral over the four sides.

Notice that your less precise statement of the Cauchy theorem, ("I know that..."), would also imply that the integral of 1/z is zero over this contour, which is false. I.e. the function needs to be analytic not just on the contour but also at points "interior to" the contour.

Oh yes, and a contour is normally oriented, but when the integral is zero, it does not matter much for the value, since zero = -zero.

Safder Aree
mathwonk said:
Let me make the statement more correct. Cauchy's theorem says that an analytic function has integral zero over a closed contour, lying in its region of analyticity, provided that contour does not wind around any point where the function is not analytic. In the case of a simple closed contour like this one, that means the function must be analytic on the contour as well as on the points interior to the contour. Does that apply?

In this extremely simple case, it might be more instructive to actually parametrize the contour and compute the integral over the four sides.

Notice that your less precise statement of the Cauchy theorem, ("I know that..."), would also imply that the integral of 1/z is zero over this contour, which is false. I.e. the function needs to be analytic not just on the contour but also at points "interior to" the contour.

That makes sense. Thank you. I'll just solve the integral.

I'm not sure if I am parametrizing this correctly, would you be able to double check where i am going wrong?

So for the path 0 -> 1.

$$z(t) = t, z'(t) = 1$$

$$\int_0^1 e^t dt$$Path 1-> 1+i
$$z(t) = 1 + it, z'(t) = 1$$
$$\int_0^1 e^{(1+it)}dt$$

Path 1+i -> i
$$z(t) = 1+i -t, z'(t) = -1$$
$$-\int_0^1 e^{1+i -t}dt$$

Path i -> 0
$$z(t) = i -t, z'(t) = -1$$
$$-\int_0^1 e^{i-t}dt$$

## 1. What is contour integration over a square?

Contour integration over a square refers to the process of evaluating a complex integral along a closed path shaped like a square in the complex plane. This method is used in complex analysis to solve integrals that are difficult or impossible to evaluate using traditional methods.

## 2. How does contour integration over a square differ from other methods of integration?

Contour integration over a square involves using a closed path in the complex plane to evaluate an integral, while traditional methods of integration involve using real numbers and functions in the real plane. This allows for the evaluation of complex integrals that would otherwise be very difficult or impossible to solve.

## 3. What is the importance of contour integration over a square in complex analysis?

Contour integration over a square is a powerful tool in complex analysis, as it allows for the evaluation of complex integrals that cannot be solved using traditional methods. This method is used in various areas of mathematics and physics, such as in the study of complex functions, differential equations, and quantum mechanics.

## 4. Are there any limitations to using contour integration over a square?

While contour integration over a square is a useful method, it does have some limitations. For example, it may not always be possible to find a suitable contour that can be used to evaluate a given integral. Additionally, the process of finding and evaluating the contour can be quite challenging and time-consuming.

## 5. Can contour integration over a square be applied to any complex integral?

No, contour integration over a square can only be applied to certain types of complex integrals. These integrals must have a closed path in the complex plane, and the function being integrated must be analytic within the region enclosed by the contour. Additionally, the contour must be chosen carefully to ensure that the integral converges.

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