This is a basic question. But it’s my first time really dealing with this topic.
When we have a integral along the real line we have:
$$\int^{b}_{a} f(x)dx$$
Which obviously has no path dependency. You simply go along from a to b.
Now when dealing with complex integrals we can have a similar integral
$$\int^{b}_{a} f(z) dz$$
Why can we not now evaluate this the same way?
Is it because z is actually split into 2 components, and so analogous to path integration of vectors, this is only meaningful when you say what path you take?
(I have currently not done group theory, or anything tremendously advanced so be easy)
When one writes $$\int_a^bf(x)dx$$ It unambiguously means along the path $\Gamma : [a,b] \mapsto [a,b]$,with $\Gamma(x) = x$ so $$\int_a^bf(x)dx = \int_\Gamma f(x)dx$$ But for complex integrals you are in a 2 dimensional vector space so multiple paths exist. So if you fix two end points $z_1$ and $z_2$ you must state how you get from $z_1$ to $z_2$. So you need something like $\Gamma :[0,1] \mapsto \mathbb{C}$ with $\Gamma(0) = z_1$ and $\Gamma(1) = z_2$ and compute $$\int_\Gamma f(z)dz = \int_{0}^{1}f(\Gamma(t))\Gamma'(t)dt$$ or if $f$ is already given by $f(t) = u(t) + iv(t)$ then you just have $$\int_a^b f(t)dt = \int_a^bu(t)dt + i\int_a^b v(t)dt$$