Partial fractions with imaginary roots: Evaluating $\mathcal{L}^{-1}\left\{\frac{1}{s^2-18s+810}\right\}$. I would like to find the Inverse Laplace transform of the following fraction:
$$\frac{1}{s^2-18s+810}$$
The problem is that the numerator has imaginary roots $9+27i$ and $9-27i$.
Can someone solve this fraction step by step, because I can't find $A$ and $B$ from imaginary roots?
 A: Instead of using partial fractions, I would start by completing the square on the denominator:
$$\frac{1}{s^2-18s+810}=\frac{1}{(s-9)^2+729} \tag{*}$$
You can now easily use result 19 on the Table of Laplace Transforms, however if you wish to derive it, I provided a small proof:


It is well known that:
  $$\mathcal{L}\{\sin(bt)\}=\frac{b}{s^2+b^2} \tag{1}$$

Proof:
By definition, we know that:
$$\mathcal{L}\{\sin(bt)\}=\int_0^{\infty} e^{-st}\sin(bt)~dt$$
Consider using integration by parts twice on the indefinite integral:
$$\begin{align}\int e^{-st}\sin(bt)~dt&=-\frac{1}{s}e^{-st}\sin(bt)+\frac{b}{s}\int e^{-st}\cos(bt)~dt\\&=-\frac{1}{s}e^{-st}\sin(bt)+\frac{b}{s}\left(-\frac{1}{s}e^{-st}\cos(bt)-\frac{b}{s} \int e^{-st}\sin(bt)~dt\right)\\&=-\frac{1}{s}e^{-st}\sin(bt)-\frac{b}{s^2}e^{-st}\cos(bt)-\frac{b^2}{s^2}\int e^{-st}\sin(bt)~dt \end{align}$$
This implies that:
$$\frac{s^2+b^2}{s^2}\cdot \int e^{-st}\sin(bt)~dt=\displaystyle -e^{-st} \left({\frac{1}{s} \sin (bt) + \frac b {s^2} \cos (bt)  }\right)$$
Therefore:
$$\begin{align}\mathcal{L}\{\sin(bt)\}&=-\frac{s^2}{s^2+b^2}\left[e^{-st}\left(\frac{1}{s}\sin(bt)+\frac{b}{s^2}\cos(bt)\right)\right]_0^{\infty}\\&=-\frac{s^2}{s^2+b^2}\left(0-1\cdot \left(\frac{1}{s}\cdot 0+\frac{b}{s^2}\cdot 1\right)\right)\\&=\frac{b}{s^2+b^2} \tag*{$\blacksquare$} \end{align}$$


And, note the first shifting theorem:
  $$\mathcal{L}\{e^{at}f(t)\}=F(s-a) \tag{2}$$

Proof:
By definition:
$$\begin{align}\mathcal{L}\{e^{at}f(t)\}&=\int_0^{\infty} e^{-st}[e^{at}f(t)]~dt\\&=\int_0^{\infty}e^{-(s-a)t}f(t)~dt\\&=F(s-a) \tag*{$\blacksquare$} \end{align}$$

Therefore, it follows from $(1)$ and $(2)$ that:
$$\mathcal{L}\{e^{at}\sin(bt)\}=\frac{b}{(s-a)^2+b^2} \tag{3}$$
We can easily apply $(3)$ to evaluate the Inverse Laplace Transform of $(*)$.
A: Since you've factored the denominator, we can set
$$\frac{1}{s^2-18s+810}=\frac{A}{s-9-27i}+\frac{B}{s-9+27i}$$
for some $A,B \in \Bbb C$. Multiplying both sides by $s^2-18s+810$ yields
$$1 = A(s-9+27i) + B(s-9-27i).$$
Now gather the $s$ components and the constants on the RHS, and equate them with the respective components on the LHS:
$$0s + 1 = (A+B)s + -9A + 27Ai - 9B - 27Bi,$$
from which we gather $0 = A+B$ and $1 = -9A + 27Ai - 9B - 27Bi$. Now solve for $A$ and $B$.
