Transfer function analisis 
I've seen this problem solved out there but I don't quite understand what it is done 
Based on the transfer function 
$$ P(s)=\frac{(s+2)}{(s+0.5)(s+4)} $$
it is asked  to graphically evaluate the unit step response.
I understand the poles and zeros are like in the picture. The step pole is in the origin of the plane, then its a zero on -2, and poles in -.5 and -4.
Then it is said to do this ( I don't understand )
$$R_{1}=\frac{2}{0.5(4)}=1; arg R_{1}=0$$
$$R_{2}=\frac{1.5}{0.5(3.5)}=0.857; arg R_{2}=-180$$
$$R_{3}=\frac{2}{4(3.5)}=0.143; arg R_{1}=-180$$
next is :
$$ y(t)=R_{1}+R_{2}e^{(-0.5t)}+R_{3}e^{(-4t)}=1-0.857e^{(-0.5t)}-0.143e^{(-4t)}$$
But how are calculated the values of $R_{1}$, what does mean that, and how the angle its calculated?

Update: The one I've done before is like this 
$$ P(s)=\frac{1}{(s+1)(s+2)} $$
The magnitude and angle of $P(jω)$ for $ω=1$
  is 

$$ P(j1)=\frac{1}{\sqrt{2}\sqrt{5}}=0.316 $$
add the angle 
$$argP(j1)=-26.6-45=-71.6$$
But I don't see how this fit in the question before.
 A: You seem as though you're under the misconception that $s$ is always an imaginary variable. This is not so; from a sophisticated viewpoint, most of the theoretical results about Laplace transforms come from treating $s$ as a complex variable and from treating the trnsfer function as a meromorphic function of that complex variable. A more accessible way to understand this viewpoint is to consider the meaning of the transfer function $H(s)$ when the input is a damped sinusoid, rather than a sinusoid, i.e. of the form $\exp((\sigma+ i\,\omega)\,t)$. The particular integral for the output of the system is still the same damped sinusoid $H(\sigma + i\,\omega)\,\exp((\sigma+ i\,\omega)\,t)$, i.e the original scaled by $H(\sigma + i\,\omega)$. You can show this by inverting the Laplace transform of the output, i.e $\frac{H(s)}{s-\sigma-i\,\omega}$, using tables. So the simple results for sinusoidal inputs generalize straightforwardly to damped sinusoids.
That being said, what one is trying to do is express the rational function $H(s)$ as a partial fraction series of the form:
$$H(s) = \sum\limits_j \left( \frac{a_{j1}}{s - s_j} + \frac{a_{j2}}{(s - s_j)^2} + \cdots + \frac{a_{j k_j}}{(s - s_j)^{k_j}} \right)\tag{1}$$
where the $s_j$ are the complex zeros of the denominator of $H(s)$ (i.e. the poles of $H(s)$) and the $k_j$ are the multiplicities of the poles. Such an expansion always exists; there is a straightforward formula for the co-efficients $a_{jk}$ and the inverse Laplace transforms for all the summed terms in the expression are known. See the "Residue Method"section of the "Partial Fraction Decomposition"Wikipedia page.
You're calculating the step response, so you're trying to find the inverse Laplace transform of
$$\frac{(s+2)}{s\,(s+0.5)(s+4)}=\frac{R_1}{s}+\frac{R_2}{s+\frac{1}{2}}+\frac{R_3}{s+4}\tag{2}$$
Here all the poles are of multiplicity 1, so the simple formula with constant numerators above holds; either use the residue method as described by the Wiki page, e.g.
$$R_2 = \lim\limits_{s\to-\frac{1}{2}}\frac{(s+2)}{s\,\left(s+\frac{1}{2}\right)(s+4)} \times \left(s+\frac{1}{2}\right) = \left.\frac{s+2}{s\,(s+4)}\right|_{s=-\frac{1}{2}}\tag{3}$$
or you can simply put all the terms on the RHS of (2) over the common denominator $s\,(s+0.5)(s+4)$ which will give you equations for the $R_j$. Once you have the $R_j$, then the RHS of (2) has inverse Laplace transform $R_1+R_2\,\exp(-t/2) + R_3\,\exp(-4\,t)$.
