Complex valued line integrals

  • Thread starter Mathman23
  • Start date
  • #1
254
0
Hi there I got a couple of question regarding the topic above

Homework Statement



(a) Given the integrals

[tex]\int \limit_{0}^{i} \frac{dz}{(1-z)^2}[/tex]
[tex]\int_{i}^{2i} (cos(z)) dz [/tex]
[tex]\int_{0}^{i\pi} e^{z} dz[/tex]

(1)write this as a Line integral on the form [tex]\int_{\gamma} f(\gamma(t)) \cdot \gamma'(t) dt[/tex] and
(2)Next find sum the of integrals using anti-derivatives.

(b)
Here am I unsure. How do I approach to calculate ?

[tex]\int_{\gamma_{n}} \frac{dz}{z}[/tex] where is [tex]\gamma_{n}:[0,2\pi] \rightarrow \mathbb{C} [/tex] is a parameter presentation of the unit circle, where [tex]n \in \mathbb{Z} - \{0\}[/tex] and which runs through

[tex]\gamma_{n}(t) = e^{itn}[/tex]


Homework Equations



Have I understood and solved (a) correctly?

What about (b) could somebody please be so kind give me a hint/(some help) :) ?

The Attempt at a Solution



attempted Solution A:

I choose [tex]\gamma(t) = t \cdot i[/tex] where [tex]t \in [0,1][/tex]

Since [tex]\gamma(t)' = i[/tex] then the solution is

[tex]\int_{\gamma} \frac{dt \cdot i}{(1-(it))^2} = \int_{0}^{1} \frac{dt \cdot i}{(1-(it))^2} = -1/2 + 1/2 \cdot i[/tex]


the finding the sum of the original integral

[tex]\int_{0}^{i} \frac{dz}{(1-(z))^2} = -1/2 + 1/2 \cdot i [/tex]

Second integral:

[tex]\int_{0}^{i\pi} e^{z} dz = \int_{\gamma} (e^{it} \cdot i) dt = \int_{0}^{\pi} (e^{it} \cdot i) dt = -2 [/tex]

finding the sum of the integral:

[tex]\int_{0}^{i\pi} e^{z} dz = -2[/tex]

(B) Attempted solution

If [tex]\int_{\gamma_{n}} \frac{dz}{z}[/tex], then to solve this integral using the definition of the line integral

then I take [tex]\int_{\gamma_{n}} \frac{dz}{z} = \int_{0}^{2\pi} \frac{1}{e^{t \cdot n \cdot i}} \cdot \frac{d}{dt}(e^{t \cdot n \cdot i}) dt = 2 \cdot n \cdot \pi \cdot i[/tex]

where [tex]n \in \mathbb{Z} - \{0\}[/tex]

Could this be the solution?

Best Regards
Fred
 
Last edited:

Answers and Replies

Related Threads on Complex valued line integrals

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
4
Views
5K
Replies
1
Views
1K
  • Last Post
Replies
23
Views
440
Replies
4
Views
3K
Replies
1
Views
904
Replies
2
Views
1K
  • Last Post
Replies
1
Views
861
Top