Q: Intuition for the complex line integral, and its relation with the line integral in $\mathbb{R}^2$. For the complex line integral, we have the following result. For $f = u + iv$, 
$$
\int_\gamma f(z)dz = \int_\gamma (u(x,y)dx - v(x,y)dy) + i \int_\gamma (v(x,y)dx+u(x,y)dy). 
$$
Can someone explain how to derive this result? I have seen that it can be derived from the mnemonic ``$dz = dx+idy$", but this is informal. In particular, I'm curious to its relation with the line integral in $\mathbb{R}^2$. This is defined, for a parameterisation $t \mapsto (x(t),y(t))$ as 
$$
\int_\gamma p(x,y)dx + q(x,y)dy = \int_a^b [p(x(t),y(t)) x'(t) + q(x(t),y(t))y'(t)] dt
$$
In my mind, the complex line integral is just a rephrasing of the line integral in $\mathbb{R}^2$ in $\mathbb{C}$, but then I would expect 
$$
\int_\gamma f(z)dz = \int_\gamma u(x,y)dx + v(x,y)dy
$$
instead of the above result, so it seems my intuition is off here. Any insights are appreciated. 

A: if you parametrize $\gamma $, i.e. $$\gamma =\{\gamma (t)=\gamma _1(t)+i\gamma _2(t)\mid t\in I\}$$
then $$\int_\gamma f(z)\,\mathrm d z=\int_I f(\gamma (z))\gamma '(t)\,\mathrm d t.$$
Now you develop in real and imaginary part, and you'll get your "informal" manipulation.
