How to understand complex (line) integral? I am taking a complex analysis class, but I have serious troubles with understanding the complex integral.
We defined it as $$ \int_{\Gamma} f(z) dz = \int_{\alpha}^{\beta} f(z(t))\dot{z}(t)dt $$,
where $z$ is the parametrization of curve $\Gamma$.
Question 1:
First let me write how I understand this formula:

*

*$z(t): \mathbb{R} \to \mathbb{C}, f(z):\mathbb{C} \to \mathbb{C}$, so $f(z(t)): \mathbb{R} \to \mathbb{C}$

*Also $\dot{z}(t): \mathbb{R} \to \mathbb{C}$, so $f(z(t))\dot{z}(t): \mathbb{R} \to \mathbb{C}$

*Therefore the integral is of the form $u(t) + i v(t)$, where $u,v: \mathbb{R} \to \mathbb{R}$.

Which means that:
$$ \int_{\Gamma} f(z) dz = \int_{\alpha}^{\beta} f(z(t))\dot{z}(t)dt = \int_{\alpha}^{\beta} (u(t) + i v(t)) dt = \int_{\alpha}^{\beta} u(t) dt + i \int_{\alpha}^{\beta} v(t) dt $$
Is this reasoning correct?
Question 2:
If so, do we always have to split the integrand on real and complex part? Because I have seen in some examples that we can directly integrate
$$ \int e^{i \varphi} d \varphi = \frac{e^{i \varphi}}{i} $$
Why is this ok? We have never derived this, yet we use it as something that is obviously true. How about $\int f(u(x) + iv(x)) dx$? Can this also be integrated as if we are integrating a real function, not a complex one?
Question 3:
Besides the definition of complex integral, we also wrote something else, that we didn't explain in any more details and I have no idea what is meant by it.
We wrote $dz = dx + idy$ and $d \overline{z} = dx - i dy$ and after that:
$$ \int_{\Gamma} f(z) dz = \int_{\Gamma} f(z)(dx+idy) = \int_{\Gamma} f(z)dx+(if(z))dy $$
and similarly for $\int d\overline{z}$.
Now, I have absolutely no idea what was meant by that. Perhaps this is just notation, but surely there is some reason or meaning behind it. Also, what does $\int d\overline{z}$ represent, geometrically / intuitively?
Thank you for any help in advance!
 A: Your reasoning in your first question seems correct. For your second question, integrals of the form $\int u + iv$ with $u,v$ real are defined as $\int u + i \int v$. Then how can you quickly calculate integrals such as the one you mention without splitting the integral into two integrals? Because the fundamental theorem of calculus also holds for complex line integrals. You can try to prove this: If $f$ is holomorphic on $\Omega$ with antiderivative $F$ and $z(t):[a,b] \to \Omega$ is a parametrization of the curve $\Gamma$, then
$$\int_{\Gamma} f(z) \, dz = F(z(b))-F(z(a)).$$
To prove this, use the chain rule; if you get stuck, see the answer to this
for example.
The basic intuition I have for complex line integrals is that they are the natural way to define integrals of $\mathbb{C} \to \mathbb{C}$ functions; if you compare the definition to the definition of integrals $\mathbb{R} \to \mathbb{R}$, it is pretty much entirely analogous: take the value of a function at the point, and multiply it by a small change in the domain of integration, and add all these changes up. In the complex case this is just a little bit more hassle since you have to parameterize your paths.
For your third question, such integrals are simply defined by
$$
\int_{\Gamma} f(z) d\bar{z} = \int_{a}^{b} f(z(t))\overline{\dot{z}(t)}\,dt.
$$
Here, instead of 'adding up' products of function values and changes in your path, you add up the products of the function value and the complex conjugate of little changes in you path; I don't have more intuition for this definition than that.
