hello I am having some trouble finding the solution to this inverse laplace transformation
$$ I(s)= \frac{6s+24}{s^2 +4s+8} $$
The solution is solved using Euler identity and partial fractions, $$ \frac{6s+24}{{(s+2-j2)(s+2+j2)}} = \frac{A}{{(s+2-j2)(s+2+j2)}} + \frac{B}{{(s+2-j2)(s+2+j2)}} $$
I understand the parital fraction portion but I dont know what to do with the solutions which are $$ A=3\sqrt2\langle45^{\circ}$$ $$B=3\sqrt2\langle-45^{\circ} $$
the final answer should be $$i(t) = 6\sqrt2e^{-2t}\cos(2t-\langle45^{\circ})$$
I have the feeling that you had a problem with partial fraction decomposition since $$\frac{6s+24}{s^2 +4s+8}=\frac{6s+24}{{(s+2-2i)(s+2+2i)}} = \frac{A}{(s+2-2i)} + \frac{B}{(s+2+2i)}$$ from which $$s (-A+B+6)+(-2-2 i) A+(2-2 i) B+24=0$$ Then $$-A+B+30=0$$ $$(-2-2 i) A+(2-2 i) B+24=0$$ from which $A=3(1-i)$, $B=3(1+i)$.
I am sure that you can take from here.