Complex valued line integrals

1. Apr 25, 2007

Mathman23

Hi there I got a couple of question regarding the topic above

1. The problem statement, all variables and given/known data

(a) Given the integrals

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

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

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

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

$$\gamma_{n}(t) = e^{itn}$$

2. Relevant equations

Have I understood and solved (a) correctly?

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

3. The attempt at a solution

attempted Solution A:

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

Since $$\gamma(t)' = i$$ then the solution is

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

the finding the sum of the original integral

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

Second integral:

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

finding the sum of the integral:

$$\int_{0}^{i\pi} e^{z} dz = -2$$

(B) Attempted solution

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

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

where $$n \in \mathbb{Z} - \{0\}$$

Could this be the solution?

Best Regards
Fred

Last edited: Apr 25, 2007