General Exponential Response Formula [ODE] I am reading MIT PDF here: http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/exponential-response/MIT18_03SCF11_s14_4btext.pdf
The 2nd question is 
and the solution is : 
My question is, how do they go from the complex form to the real form? Re(z), what is this operation. 
Also, if the function had been a sin(t) instead of cos(t) in the ODE, then I would take Im(z), how do I do this operation aswell?
An example where that might apply is let's say x''+5'x+6x = sin(5t)
 A: $Re(x)$ gives you the real part: $Re(x) = Re(a + bi) = cos(a)$
$Im(x)$ gives you the imagniary part: $Im(x) = Re(a + bi) = sin(b)$
In your example: $x_P  = Re(z_p) = Re(\frac{te^{-t}( cos(t) + i sin(t) )}{2i}) = Re(\frac{te^{-t}cos(t)}{2i} +  \frac{te^{-t}sin(t)}{2}) = Re(-\frac{ite^{-t}cos(t)}{2} +  \frac{te^{-t}sin(t)}{2}) =  \frac{te^{-t}sin(t)}{2}$
$Im(z_p) = Im(-\frac{ite^{-t}cos(t)}{2} +  \frac{te^{-t}sin(t)}{2}) =  -\frac{te^{-t}cos(t)}{2}$.
A: let us take the example you have. $$x'' + 5x' + 6x = \sin 5t.$$ then $$x = e^{\lambda t}  $$ is a solution of the homogenous equation if $$\lambda^2 + 5 \lambda + 6 = 0 \to \lambda = \frac{-5 \pm \sqrt{1}}{2}=-2, -3$$ unfortunately, this equation has no complex roots.  
let us try instead $$x'' + 4x' + 6x = \sin 5t.$$ the char equation is 
$$\lambda^2 + 4 \lambda + 6 = 0 \to \lambda = \frac{-4 \pm \sqrt{-8}}{2}=-2\pm\sqrt 2i$$ 
you can verify that both $$ x = e^{-2t}\cos\sqrt 2 t,\, x = e^{-2t}\cos\sqrt 2 t  $$
satisfy the homogeneous equation.
these two solution can be thought as coming from the real and imaginary parts of 
$$e^{-2t + i\sqrt 2t} = e^{-2t}\left( \cos \sqrt 2 t + i \sin \sqrt 2 t\right).$$ i have used the eulers formula $e^{it} = \cos t + i \sin t.$
