This is an interesting complex analysis problem; The figure on the bottom left is what is being referred to,Fig7-10.
First, lets take a look at the complex line integral.
What is the geometry of the complex line integral?
If we look here enter link description here
The real line integral is a path, but then you make a 3d figure, and it is the area under the 3d shape.
What about for complex integral?
And
How is in the solution:
$$\int_{BDEFG} = \int_{0}^{2\pi} \frac{(Re^{i\theta})^{p-1}iRe^{i\theta} d\theta}{1 + Re^{i\theta}}$$
How is the integral around the $BDEFG$, the same as the area from 0 to $2\pi$
Thanks.
1) I don't think there is a simple geometric interpretation of a contour integral. The picture you've linked talks about functions $f: \mathbb{R}^n\to \mathbb{R}$. The so called scalar fields. While you are working with $f: \mathbb{R}^2\to \mathbb{R}^2$.
If you would like to dive into geometric approaches to complex analysis I would advise "Visual Complex Analysis" by Needham. (If you got the time to dig through it)
2) Take the parameterization $BDEFG\leftrightarrow z(\theta) = Re^{i\theta}$ with $\theta$ from $0 \to 2\pi$ and $z'(\theta) = iRe^{i\theta}$.
$$\int\limits_{BDEFG} \frac{z^{p-1}}{1+z} \operatorname dz = \int_0^{2\pi} \frac{(Re^{i\theta})^{p-1}}{1+Re^{i\theta}}\cdot iRe^{i\theta} \operatorname d \theta$$