Contour integration by parametisation

In summary, the problem involves determining the contour integral ∫C dz f(z), where f(z) = z^n and C is a line segment from z = 0 to z = 2, and then from z = 2 to z = 2 + i. The solution involves splitting the contour into two sub-contours and using the equation ∫ϒ f(z) dz = ∫ab dt f(ϒ(t))ϒ'(t) to integrate over both and sum the results. The resulting answer may be complex, depending on the value of n.
  • #1
Poirot
94
2

Homework Statement


Determine, by explicit parameterisation,

C dz f(z), where f(z) = z^n , n ∈ Z, n > 0
and C is the line segment from z = 0 to z = 2 together with the line segment from z = 2 to z = 2 + i.

Homework Equations


ϒ f(z) dz = ∫ab dt f(ϒ(t))ϒ'(t) where ϒ'(t)≈dz/dt

The Attempt at a Solution


Splitting C into two sub contours ϒ1 = {z=x, x ∈ [0 , 2]} and ϒ2 = {z=y, y ∈ [2 , 2+i]}. Then using the equation above to integrate over both and sum. This gave and answer which I really can't gauge whether is correct or not.

(2+i)(n+1)/(n+1)

But my lecturer gave us a rough check guide and said that if what you start with is a real number then your answer should be real also. Thus am skeptical of my answer.

Any help on the method, answer or general help with this would be greatly appreciated.

Thanks!
 
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  • #2
Poirot said:

Homework Statement


Determine, by explicit parameterisation,

C dz f(z), where f(z) = z^n , n ∈ Z, n > 0
and C is the line segment from z = 0 to z = 2 together with the line segment from z = 2 to z = 2 + i.

Homework Equations


ϒ f(z) dz = ∫ab dt f(ϒ(t))ϒ'(t) where ϒ'(t)≈dz/dt

The Attempt at a Solution


Splitting C into two sub contours ϒ1 = {z=x, x ∈ [0 , 2]} and ϒ2 = {z=y, y ∈ [2 , 2+i]}. Then using the equation above to integrate over both and sum. This gave and answer which I really can't gauge whether is correct or not.

(2+i)(n+1)/(n+1)

But my lecturer gave us a rough check guide and said that if what you start with is a real number then your answer should be real also. Thus am skeptical of my answer.

Any help on the method, answer or general help with this would be greatly appreciated.

Thanks!
It's not clear what your lecturer was talking about, or if you have fully understood what he said, but I have never heard of such a "rough check" on open contour integrals.

In fact, a similar problem is contained in this article (pp. 4-5) where f(z) = z2 and the contour is dog-legged like the one for your problem:

https://math.dartmouth.edu/archive/m43s12/public_html/notes/class6.pdf

The calculation starts out with a real result on C1 and gives a complex result on C2 with a net result which is complex. :wink:
 

1. What is contour integration by parametisation?

Contour integration by parametisation is a method used in complex analysis to evaluate integrals along a contour (or path) in the complex plane. It involves parameterizing the contour with a complex variable, which allows for the use of techniques from calculus such as the chain rule and substitution.

2. Why is contour integration by parametisation useful?

Contour integration by parametisation is useful because it allows for the evaluation of integrals that cannot be solved using traditional methods. It also allows for the evaluation of integrals over closed contours, which can be used to calculate the values of complex functions at certain points.

3. How is contour integration by parametisation different from other methods of integration?

Unlike traditional methods of integration, contour integration by parametisation deals with complex numbers and uses techniques from complex analysis. It also involves integrating over a contour rather than a real interval, which requires the use of parametrisation.

4. What are some applications of contour integration by parametisation?

Contour integration by parametisation is used in many areas of mathematics and physics, including complex analysis, signal processing, and quantum mechanics. It is also used in engineering applications, such as in circuit analysis and control theory.

5. Are there any limitations to contour integration by parametisation?

Contour integration by parametisation can be limited by the complexity of the contour and the function being integrated. If the contour is too complicated or the function has poles or branch points, the parametrisation may become difficult or impossible. In these cases, other methods of integration may be more suitable.

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