When exactly is a path-Integral equal to 0 and when to apply Cauchy's Integral formula for derivatives

Question

So I am just looking at some exercise to prepare myself for the exam and just seem confused when I can apply these Theorems:

a) Let $A = 1 + i, B = −1 + i, C = −1 − i, D = 1 − i$ and γ be the path [A, B, C, D, A], that is the contour of the square ABCD.

Compute the path Integrals $\int_{γ} \frac{e^{z}-e^{-z}}{e^{z}-4}$ and $\int_{γ} \frac{e^{z}}{{z}^4}$

So to the first Integral I know, that the function $f(z) = \frac{e^{z}-e^{-z}}{e^{z}-4}$ is holomorphic and thus continous on the entire Complex plane and γ is a closed path.

One of my Theorem only states, that if the domain is star-shaped and f is holomorphic on that domain, then f has an antiderivative. And if this is true and we are integrating over a closed curve, the Integral must be equal 0.

So why can I apply this to this example, or is there an easier way of doing it?

The second Integral can be solved with Cauchy's Integral formula for derivatives, but I am also not sure why I can apply it here?

Some explanation would help me a lot :)

Answer 1

For $$\int_{γ} \frac{e^{z}-e^{-z}}{e^{z}-4}$$ you may use the fact that the integral of a G-holomorphic function on a closed curve contained in an open region G is $0$.

For the second one $$ \int_{γ} \frac{e^{z}}{{z}^4} $$ you may use the Cauchy's Integral formula for third derivatives because the function $e^z$ is an entire function.

Answer 2

In the case of the first integral, you can simply use Cauchy's theorem in order to prove that it is equal to $0$, since the path is null-homotopic in $\mathbb{C}\setminus\{z\in\mathbb{C}\,|\,e^z=4\}=\mathbb{C}\setminus\{\log 4+2k\pi i\,|\,k\in\mathbb Z\}$.

For the other integral, just use the residue theorem. It tells you that$$\int_\gamma\frac{e^z}{z^4}\,\mathrm dz=2\pi i\operatorname{res}_{z=0}\left(\frac{e^z}{z^4}\right)=\frac{\pi i}3.$$