I have a function which I'd like to perform partial fraction decomposition on, to allow easier inverse laplace transform.
$$F(s) = \frac{1}{s(s^2+140s+10^4)}$$
I begin with finding the poles
$$s = 0, s = -70 \pm j \cdot 10\sqrt{51}$$
To which I then try putting $F(s)$ in this form:
$$F(s) = \frac{A}{s} + \frac{B}{s-70+j\cdot 10\sqrt{51}} + \frac{B^*}{s-70-j\cdot 10\sqrt{51}}$$
Because one unknown ($B^*$) is just the complex conjugate of $B$, I only need to find out what $A$, and $B$ is.
$$A = F(0) = 10^{-4}$$
$$B = F(-70+j\cdot10\sqrt{51}) \approx 6.31514\cdot10^9 + j\cdot 6.44274\cdot10^9$$
Where the last step was done in Mathematica.
Answer, according to several other people, is supposed to be $B = -70 + j\cdot71.41$, which looks a lot nicer, but I'm not sure HOW they got to that answer.
Note that the partial fraction expansion of $F(s)$ is
$$F(s)=\frac{A}{s}+\frac{B}{s-s_0}+\frac{B^*}{s-s_0^*}$$
where $s_0= -70+i10\sqrt{51}$ and $s_0^*=s_0= -70-i10\sqrt{51}$ are the complex-conjugate roots of $s^2+140s+10^4$.
Note that we have
$$\begin{align} A&=\lim_{s\to 0}sF(s)\\\\ &=10^{-4}\\\\ \end{align}$$
and
$$\begin{align} B&=\lim_{s\to s_0}(s-s_0)F(s)\\\\ &=\frac{1}{s_0(s_0-s_0^*)}\\\\ &=\frac{-70-i10\sqrt{51}}{i20\sqrt{51}10^4 }\\\\ &=\frac{-51+i7\sqrt{51}}{1020000} \end{align}$$