Inverse Laplace Transform via Circuit Analysis [HELP] Inverse Laplace Transform $\frac{1}{s^2 + \sqrt{2}s + 1}$
so what I did it changed the denominator to complete the square format which is $\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$, then I can solve for $s$, it will make it as
$$
\left(\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i\right)
\left(\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i\right)
$$
So now, to the sheet of paper is to do Partial Fraction Decomposition of this which is absurd to me because of complex roots it has:
$$
\frac{1}{s^2 + s\sqrt{2} + 1}
= \frac{1}{\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}}
$$
Partial Fraction of Complex root will be
$$
\frac{K}{\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i}
 + \frac{K^*}{\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i}
$$
to follow the formula sheet.
which I got my K = -$i\frac{\sqrt{2}}{2}$ and $K^*$ = $i\frac{\sqrt{2}}{2}$
the problem I get is magnitude and $\theta$ is undefined it makes no sense at all.


Once we complete the square, we can use the sine formula and Frequency Shift Theorem to evaluate the inverse transform:
If we accept that
$$\mathcal{L}(\sin(at)) = \frac{a}{s^2+a^2}$$
and
$$\mathcal{L}(e^{ct}f(t)) = F(s-c)$$
where $F(s) = \mathcal{L}(f(t))$, we can take our original fraction:
$\begin{align}
\mathcal{L}^{-1}(\frac{1}{s^2+\sqrt{2}s+1}) & = \mathcal{L}^{-1}(\frac{1}{(s+\frac{1}{\sqrt{2}})^2+1/2})\\
& = \mathcal{L}^{-1}(\sqrt{2}\frac{\frac{1}{\sqrt{2}}}{(s+\frac{1}{\sqrt{2}})^2+1/2})\\
& = \sqrt{2}*\exp{\frac{-t}{\sqrt{2}}}*\sin(\frac{t}{\sqrt{2}})
\end{align}$
In that last step, we combined the two formulae above, as our fraction was in the form of $\mathcal{L}(\sin(at))$, but shifted by $c = \frac{-1}{\sqrt{2}}$, creating the '$\exp{\frac{-t}{\sqrt{2}}}$' term in the final answer.
Were you to continue the partial fraction decomposition method directly, you would end up with two exponentials terms that you could manipulate into the same answer above using the identity:
$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$
