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.

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.
