In an exercise, a partial fraction expansion has to be done. I have no problem with that, but one of the last steps includes a simplification as follows:
\begin{equation*} \left( -\frac 12 - \frac 16 i \right) \left(\frac{1}{s+1+3i}\right) +\left( -\frac 12 +\frac 16 i \right) \left(\frac{1}{s+1-3i}\right) \\ = -\frac{s+1}{(s+1)^2+3^2}-\frac 13\left( \frac{3}{(s+1)^2+3^2} \right). \end{equation*}
My problem is that I do not understand the steps in between these two results. Could anyone explain how this is done? Any help is appreciated!
The first step is usually to realify(?) the denominators of your complex fraction. To do this, multiply the first term on the LHS by $1=\frac{s+1−3i}{s+1−3i}$. Multiply the second by $1=\frac{s+1+3i}{s+1+3i}$. These are just the complex conjugates of the denominators. Also, once you obtain the RHS, notice that you can simplify it further. So let's try:
$$\left( -\frac 12 - \frac 16 i \right) \left(\frac{1}{s+1+3i}\right) +\left( -\frac 12 +\frac 16 i \right) \left(\frac{1}{s+1-3i}\right) \\ = -\frac 16\left(\frac{3+i}{s+1+3i}\right) +\frac 16\left(\frac{-3+i}{s+1-3i}\right) \\ = -\frac 16\left(\frac{3+i}{s+1+3i}\right)\left(\frac{s+1-3i}{s+1-3i}\right) +\frac 16\left(\frac{-3+i}{s+1-3i}\right)\left(\frac{s+1+3i}{s+1+3i}\right) \\ = -\frac 16\left(\frac{[3(s+1)+3]+[-9+(s+1)]i}{(s+1)^2+3^2}\right) +\frac 16\left(\frac{[-3(s+1)-3]+[-9+(s+1)]i}{(s+1)^2+3^2}\right) \\ = \left(\frac {-(s+1)-1}{(s+1)^2+3^2}\right) \\ = -\left(\frac {s+2}{(s+1)^2+3^2}\right)$$