Find that $Ze^{zt} + \overline{Z}e^{\overline{z}t} = 2Re(Ze^{zt})$ I'm currently learning about oscillations by mass-spring systems with damping in one dimension. We start with: \[ m\frac{d^2x}{dt^2}  +  b\frac{dx}{dt} + kx = 0 \]
Which we transform to:
\[ \frac{d^2x}{dt^2}  +  2\gamma \frac{dx}{dt} + \omega_0^2x = 0 \]
Where $\gamma = \frac{b}{2m}$ and $\omega_0 = \sqrt{\frac{k}{m}}$. Which has the characteristic equation:
\[ z^2 + 2\gamma z + \omega_0^2 = 0 \]
If we now look at when $ D < 0 $, we find that $b^2 < 4km$ which is an underdamped oscillation. We have \[ z_1 = -\gamma +i\sqrt{\omega_0^2-\gamma^2} = -\gamma + i\omega \]
\[  z_2 = \overline{z_1} = -\gamma - i\omega \]
now my textbook does the following for x(t):
\[ x(t) = Ze^{z_1t} + \overline{Z}e^{z_2t} = 2Re(Ze^{z_1t}) \]
Where Z is a complex number. I understand the solution to the differential equation, but I can't seem to prove that the 2nd part is equal to the 3rd part. I've tried writing it out, but I get stuck on trying to find the real part of an exponential so if anyone could help me it would be greatly appreciated!
 A: Note that $\overline{e^{a+ib}}=\overline{e^a}\overline{e^{ib}}=e^{a}e^{-ib}=e^{a-ib}=e^{\overline{a+ib}}$. Since $z_2=\overline{z_1}$, we have
$$ Ze^{z_1t}+\overline{Z}e^{z_2t}=Ze^{z_1t}+\overline{Ze^{z_1t}}=2{\rm Re}\left(Ze^{z_1t}\right) $$
because $z+\overline{z}=2{\rm Re}(z)$ for all $z\in\mathbb{C}$.
A: All you need is that $\overline{e^z} = e^{\overline z}$, since then
$$
Ze^{zt} + \overline Ze^{\overline zt} = Ze^{zt} + \overline{Ze^{zt}} = 2\operatorname{Re}(Ze^{zt}).
$$
And indeed,
\begin{align*}
\overline{e^z}
&= \overline{e^{a+ib}} = \overline{e^ae^{ib}} = e^a\overline{\cos(b)+i\sin(b)}\\
&= e^a(\cos(b)-i\sin(b)) = e^a(\cos(-b)+i\sin(-b))\\
&= e^ae^{-ib} = e^{a-ib} = e^{\overline z}.
\end{align*}
A: In the case of the complex exponential it is true that
$\overline{e^z}=e^{\overline{z}}$
The decomposition of the exponential into a real part and an imaginary part is
Let $z=a+bi$, with $a$ and $b$ real numbers.
$e^z=e^{a+bi}=e^a.e^{bi}=e^a(\cos(b)+i\,\sin(b))=e^a\cos(b)+i\,e^a\sin(b)$
