Inverse Laplace Through Complex Roots I have been asked to apply inverse laplace to this:
$$ \frac{(4s+5)}{s^2 + 5s +18.5} $$
What I have done is;
I found the roots of denominator which are : $$ (-5-7i)/2 $$ and $$ (-5+7i)/2 $$
Then I factorized the denominator as :
$$ \frac{(4s+5)}{(s+\frac{(5+7i)}{2})(s + \frac{(5-7i)}{2})}  $$
Then i split this fraction to sum of two different fractions through;
$$ \frac{(A)}{(s+\frac{(5+7i)}{2})}  $$
$$ \frac{(B)}{(s+\frac{(5-7i)}{2})}  $$
Then I found A and B as ; $$ A = 2 + \frac{35i}{49} $$
$$ B = 2 - \frac{35i}{49} $$
At the and , Inverse Laplace took this form ;
$$ (2 + \frac{35i}{49})(\mathcal{L}^{-1}(\frac{1}{s+\frac{5+7i}{2}})) + (2 - \frac{35i}{49})(\mathcal{L}^{-1}(\frac{1}{s+\frac{5-7i}{2}}))$$
When I took the inverse laplace of these, the result was;
$$ (2 + \frac{35i}{49})( (e)^{(-3.5i - 2.5)t} ) +  (2 - \frac{35i}{49}) ( (e)^{(3.5i-2.5)t} ) $$
I verified this result from Wolfram Alpha and Mathematica. But my guest professor insists this is not the solution and he gave me 0 points. He insists the solution includes cosines and sines. I explained him if he uses Euler Identity on these exponents the result will become his result but he refuses and says only way to solve this is to use Laplace tables.
I do agree making the denominator a full square and use Laplace table is the easier and cleaner solution. But isn't this also a solution? Thanks.
 A: If these two functions are the same, I don't see why your professor is insisting on using tables as the only solution, as you came up with another one which is equivalent and gives the same function.  That being said, I don't see what is wrong with your solution (I didn't spot an error in what you did).
To follow the method your professor is demanding you follow, try rewriting this to make it look a bit more like the Laplace transform of $\sin(t)$, $\cos(t)$.
\begin{align*}
\frac{4s+5}{s^2+5s + 18.5} &= \frac{4s+5}{s^2+5s+6.25+12.25}\\
&= \frac{4s+10-5}{(s+2.5)^2+(3.5)^2}\\
&= \frac{4s+10}{(s+2.5)^2+(3.5)^2} - \frac{5}{(s+2.5)^2+(3.5)^2}\\
&= 4\left(\frac{s+2.5}{(s+2.5)^2+(3.5)^2}\right) - \frac{5}{3.5}\left(\frac{3.5}{(s+2.5)^2+(3.5)^2}\right)\\
&= 4\left(\frac{s+2.5}{(s+2.5)^2+(3.5)^2}\right) - \frac{10}{7}\left(\frac{3.5}{(s+2.5)^2+(3.5)^2}\right)
\end{align*}
So the inverse Laplace transform of $\frac{4s+5}{s^2+5s+18.5}$ is
$$4e^{-2.5t}\cos(3.5t) - \frac{10}{7}e^{-2.5t}\sin(3.5t)$$
So the only question left is: does the function given simplify to this?
\begin{align*}
&\quad \left(2+\frac{35}{49}i\right)e^{-2.5t}(\cos(3.5t) - i\sin(3.5t)) + \left(2-\frac{35}{49}i\right)e^{-2.5t}(\cos(3.5t)+i\sin(3.5t))\\
&= \left(2+\frac{5}{7}i\right)e^{-2.5t}(\cos(3.5t) - i\sin(3.5t)) + \left(2-\frac{5}{7}i\right)e^{-2.5t}(\cos(3.5t)+i\sin(3.5t))\\
&= 4e^{-2.5t}\cos(3.5t) + \frac{10}{7}e^{-2.5t}\sin(3.5t)
\end{align*}
Considering the complex exponential gives a very clean way of deriving many of the identities given in that table, your solution makes more sense than the one I gave (the "correct" solution, according to your professor).  The only issue I see at all is perhaps that this function isn't (upon first glance) obviously real-valued.  But these are the same function, and if you can argue it this simply, it shouldn't be an issue.
A: I checked carefully your calculations and they are perfectly correct ! Congratulations. May be your professor would prefer $\frac{35}{49}$ to be replaced by $\frac{5}{7}$ !!
You are totally correct with the fact that sines and cosines would be obtained using Euler's identity. So, factor $e^{-\frac {5}{2}t}$ and give $$\frac{2}{7} e^{-5 t/2} \left(14 \cos \left(\frac{7 t}{2}\right)-5 \sin \left(\frac{7
   t}{2}\right)\right)$$ in which everything is real. Probably, your professor does not enjoy complex numbers in real valued functions.
