I understand what integration of real numbers is. I know how the definition of it is made.
I have trouble in understanding how it works for complex numbers.
I am referring to the notes here: http://people.math.gatech.edu/~cain/winter99/ch4.pdf.
I understand till how the Reimann sum (page 2) is calculated because it is almost the same as reals. There after, I do not understand a thing.
Please elucidate what it is.
On
Some explanations on the examples the OP is interested in, namely those in
http://people.math.gatech.edu/~cain/winter99/ch4.pdf
at pag. 4.3-4.4.
We want to integrate the function $f(x,y)=u(x,y)+iv(x,y)=(x^2+y)+i(xy)$ along different paths. Using the notation $z=x+iy$, we have
$$f(z)=\operatorname{Re}(z)^2+\operatorname{Im}(z)+i(\operatorname{Re}(z)\operatorname{Im}(z)).$$
Remark: the function $f$ is not differentiable at $z$ because it does not satisy the Cauchy Riemann equations.
The locus of points described by the path $t\mapsto \gamma_1(t)$ is a parabola in the complex plane. In fact, by definition
$$\operatorname{Im}(\gamma_1(t))=\operatorname{Re}(\gamma_1(t))^2,$$ for all $t\in [0,1]$.
$$\operatorname{Re}(\gamma_2(t))=1$$
with $t\in [0,1]$. Their imaginary part $$\operatorname{Im}(\gamma_2(t))=t$$ increases linearly from the minimum $\operatorname{Im}(\gamma_2(0))=0$ to the maximum $\operatorname{Im}(\gamma_2(1))=1$.
EDIT. Extra example for the OP
Let $\gamma(t):=t+i\sqrt{1-t^2}$ for $t\in[0,1]$. As
$$\operatorname{Re}(\gamma(t))^2+\operatorname{Im}(\gamma(t))^2= t^2+(1-t^2)=1,$$
then the locus of points $\gamma(t)$ in the complex plane is a 1/4 of a circle of radius $1$, with initial point $\gamma(0)=0+i$ and final point $\gamma(1)=1$ (direction:clockwise).
Let us introduce the simplified notation $x(t):=\operatorname{Re}(\gamma(t))$ and $y(t):=\operatorname{Im}(\gamma(t))$
If $f(x,y)=u(x,y)+iv(x,y)=(x^2+y)+ixy$, then
$$f(\gamma(t)):=u(x(t),y(t))+i v(x(t),v(t))=(x(t)^2+y(t))+ix(t)y(t)= (t^2+\sqrt{1-t^2})+i t\sqrt{1-t^2}.$$
A comment on terminology. "Locus" (http://en.wikipedia.org/wiki/Locus_(mathematics)) denotes a set of points satisfying certain conditions. When I write "the locus of points $\gamma(t)$..." I mean the set of all points in the complex plane which are of the form
$$\gamma(t)=x(t)+iy(t).$$
In other words, here "locus" means the image in the complex plane of the map (called also curve) $t\mapsto \gamma(t)$.
As far as contours go, a rough (and probably poor) analogy is suppose you want to integrate the height of a hill. You would lay out a tape measure and measure the height at each point as you walk up the hill - except you can walk up the hill in lots of different ways. You can walk up in a straight line from east to west, or maybe a curve from south to northwest, etc. Maybe you only want to walk AROUND the top of the hill. You can think of the complex function that you're integrating as the hill (except it has a complex height instead of a real height), and the line or curve you're integrating along as the path in the complex plane.
You'll get different measurements depending on where the start and end points are, and how well-behaved the curve is. If the curve is very well-behaved, there's a nifty property that it doesn't matter what the curve is, the integral will be the same as long as the start and end are the same. You could start at point $a=1$, head all the way out to point $c=10^{1000}$ then come back to point $b=2$, and as long as your path is continuous, you will get the same answer as if your curve went straight from $a$ to $b$.
From there, you can imagine that for these well-behaved functions, the curve from $a$ to $a$ - a closed path, would be the same as the integral from $a$ to $a$ - 0, and you would be right.
I'm glossing over a lot here, but this is a very high-level view of how you can visualize complex integration.