Solving and manipulating the damped oscillator differential equation

[Hamiltonian]

the differential equation that describes a damped Harmonic oscillator is:
$$\ddot x + 2\gamma \dot x + {\omega}^2x = 0$$ where $\gamma$ and $\omega$ are constants.
we can solve this homogeneous linear differential equation by guessing $x(t) = Ae^{\alpha t}$
from which we get the condition:
$$\alpha = -\gamma \pm ({\gamma}^2 - {\omega}^2)^{1/2}$$

we can then say that the most general solution for $x(t)$ is:
$$x(t) = e^{-\gamma t}(Ae^{\Omega t} + Be^{-\Omega t})$$ where $\Omega = ({\gamma}^2 - {\omega}^2)^{1/2}$

since I am trying to derive $x(t)$ for under damping i.e. $\gamma < \omega$ or ${\Omega}^2 < 0$
say $w' = ({\omega}^2 - {\gamma }^2 )^{1/2}$
then we have $\Omega = (-1)^{1/2}w' = Iw'$

$$x(t) = e^{-\gamma t}(Ae^{Iw't} + Be^{-Iw't})$$

$x(t)$ has to be real as it represents displacement.
My book thus concludes that $$x(t) = Ce^{-\gamma t}cos(w't + \phi) ->(1)$$
it suggests we can get this by using $e^{i\theta} = cos\theta + isin\theta$ and it also states that if $x(t)$ is to be real then $A* = B$ where $A*$ is the complex conjugate of $A$(I don't see why this has to be true). I need help arriving at eqn(1).

 

[anuttarasammyak]

As you assume $A=B^*=a e^{i\phi}$. Your result goes
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
You see $C=4a$

EDIT Correction
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega't+\phi)}+e^{-i(\omega't+\phi)}]=e^{-\gamma t} 4a \cos(\omega't+\phi)$$

 

[Hamiltonian]

As you assume $A=B^*=a e^{i\phi}$. Your result goes
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
You see $C=4a$

i don't understand how they are converting the exponential form(i.e. $x(t) = e^{-\gamma t}(Ae^{\Omega t} + Be^{-\Omega t})$) to the nice trigonometric form by applying Euler's formula.

 

[Hamiltonian]

As you assume $A=B^*=a e^{i\phi}$. Your result goes
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
You see $C=4a$

EDIT Correction
$$x(t)=e^{-\gamma t} 2a [e^{i(\omega't+\phi)}+e^{-i(\omega't+\phi)}]=e^{-\gamma t} 4a \cos(\omega't+\phi)$$

$$x(t)=e^{-\gamma t} a [e^{i(\omega'+\phi)t}+e^{-i(\omega'+\phi)t}]$$
so, should'nt $C = 2a$?

 

[Hamiltonian]

I tried doing the same manipulation for the differential equation describing simple harmonic motion:
$$\ddot x + \omega^2 x = 0$$
after guessing the solution to be of form $x(t) = Ae^{\alpha t}$ we get
$$x(t) = Ae^{i\omega t} + Be^{-i\omega t}$$
now again from here I want to write $x(t)$ in terms of a trigonometric function:
since $x(t)$ has to be real $A* = B = ae^{i\phi}$
$$x = ae^{i(wt - \phi)} + ae^{-i(wt - \phi)}$$
$$x = Ccos(wt - \phi)$$ so $C = 2a$
is this correct?

 

[anuttarasammyak]

You may remember
$$cos \phi = \frac{e^{i\phi}+e^{-i\phi}}{2}$$

 

[vela]

It also states that if x(t) is to be real then A∗=B where A∗ is the complex conjugate of A(I don't see why this has to be true).

Using Euler's formula, you get
$$x(t) = A e^{i\omega t}+B e^{-i\omega t} = (A+B) \cos\omega t + i(A-B)\sin\omega t.$$ Let $A = a+bi$ and $B = c+di$. For $x(t)$ to be real, $(A+B)$ has to be real and $(A-B)$ has to be imaginary, so you get
\begin{align*}
A+B &= (a+c)+i(b+d) \quad \Rightarrow \quad d = -b \\
A-B &= (a-c) + i(b-d) \quad \Rightarrow \quad c = a
\end{align*} So you have $B = a-bi = A^*$.

The solution is now $x(t) = c_1 \cos \omega t + c_2 \sin \omega t$ where $c_1$ and $c_2$ are real. Try expanding $C \cos(\omega t - \phi)$ using the angle-addition identity and compare the two forms of the solution.