How Do You Solve Complex Valued Line Integrals?

[Mathman23]

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

Homework Statement

(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

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

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 $$\gamma(t) = t \cdot i$$ 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

 

[mighty2000]

Hi Fred,

Your attempted solutions for (a) and (b) both seem to be correct. For (a), you correctly used the definition of a line integral to rewrite the given integrals. For (b), you used the parameterization of the unit circle to rewrite the integral and then applied the definition of a line integral.

To approach (b), you could also use Cauchy's Integral Formula, which states that for a function f(z) that is analytic in a simply connected region D, and a closed curve \gamma in D, we have:

\int_{\gamma} \frac{f(z)}{z-z_0} dz = 2 \pi i \cdot f(z_0)

where z_0 is any point inside the region D enclosed by \gamma. In this case, we have f(z) = 1 and z_0 = 0, so the integral becomes:

\int_{\gamma_n} \frac{1}{z} dz = 2 \pi i

This approach may be easier for some people, but your solution using the definition of a line integral is also correct.

Hope this helps!