Okay it's 6 steps and if you do it enough times it will be down to only 5 steps.
Just so all of this actually makes sense, we will be using the following example:
$$ f(z) = \frac{1}{z} $$
Above equation is to be integrated along the circle|z|=R, starting and finishing on the route.
Before getting into the sateps lets make use we know the basics
Basic Definition
An imaginary number is conventionally denoted as
$$ z = x + \iota y $$
and a function is denoted as $$ f(z) = u(x, y) + \iota v(x, y) $$
here `u` and `v` are functions of `x` and `y`, and that `i` thing is iota also known as root of -1.
1) Path
Write the path in terms of only one variable, conventionally `t`. So, here we can write 'z' as:
$$ z = R \cos t + \iota R \sin t $$
from this we can directly compare and get `x` and `y`
$$ x = R \cos t $$
$$ y = R \sin t $$
That was all step 1 was.
2) Function to x y
Now we will get function which is in terms of 'u' and 'v' into terms of 'x' and 'y'.
$$ f(z) = \frac{1}{z} $$
$$ f(z) = \frac{1}{x + \iota y} $$
$$ f(z) = \frac{x-\iota y}{x^2 + y^2} $$
From this we can take out values of 'u' and 'v' as
$$ u = \frac{x}{x^2+y^2};v = -\frac{y}{x^2+y^2} $$
3) uv to t
At this step substitute the earlier values of 'u' and 'v' we have in terms of 'x' and 'y', to have it in terms to 't'.
$$ u = \frac{\cos t} {R} $$
$$ v = -\frac{\sin t} {R} $$
We are almost done now.
4) Piecing it all together
Now plug them into this equation you can do the following.
$$ \int{f(z)} = \int{u\frac{dx}{dt}dt} - \int{v\frac{dy}{dt}dt} + \iota\int{u\frac{dy}{dt}dt} +\iota\int{u\frac{dx}{dt}dt} $$
5) Precaution
In the very likely case you forget the equation:
You can get it directly from this:
$$ \int{f(z)dz} = \int{(u+\iota v)(dx + \iota dy)} $$
Just multiply those two brackets in the L.H.S and you'll be at the equation above it in 2 lines.