Trigonometric functions and complex numbers

I solving the inverse Laplace transform using the method of Heaviside. This is part of the problem: I understand the division between complex numbers and that $e^{it} = Cos(t) + iSin(t)$, but I got stuck in the third row.I developed the product but I don't quite understand how it came to $-Cos(t)+ \frac{1}{2}Sin(t)-Cos(t)+ \frac{1}{2}Sin(t)$

I'm not entirely sure which step is confusing you. So starting from:

$$=\cfrac{4}{2} e^t + \cfrac{3i+1}{-2-2i} e^{it} + \cfrac{-3i+1}{-2+2i}e^{-it}$$

As you noted, the next step is simply just the division of complex numbers

$$\cfrac{3i+1}{-2-2i}=\cfrac{3i+1}{-2-2i}\cfrac{-2+2i}{-2+2i}=\cfrac{-4i-8}{8}=(-1-1/2 i)$$

$$\cfrac{-3i+1}{-2+2i}=\cfrac{-3i+1}{-2+2i}\cfrac{-2-2i}{2-2i}=\cfrac{4i-8}{8}=(-1+1/2 i)$$

And the fact that $e^{it} = \cos{t}+i\sin{t}$, $e^{-it}=\cos{t}-i\sin{t}$.

Thus \begin{aligned} &=2e^t + (-1-1/2i)(\cos{t} + i\sin{t}) + (-1+1/2i)(\cos{t}-i\sin{t})\\ &=2e^t - \cos{t}-i\sin{t} -1/2 i \cos{t} +1/2\sin{t} -\cos{t}+i\sin{t} +1/2i \cos{t} +1/2\sin{t}\\ &=2e^t-2\cos{t}+\sin{t} \end{aligned}

The last step simplifies a lot because we note that all the $a i \sin{t}$ and $b i \cos{t}$ terms cancel out.

• Oh such a dumb mistake! Now I see haha Thanks! – Sandra Torres Jun 5 '16 at 20:26