<s>[INST] Using Eulers formula I am trying to figure out how 
\begin{equation*}
e^{i(-1+i\sqrt{3})}=e^{-\sqrt{3}} (cos(1)-i sin(1))??
\end{equation*}
I know that Euler's formula states that 
\begin{equation*}
e^{ix} = \cos(x) + i \sin(x)
\end{equation*}
but surely that would mean in this case $x=-1+i\sqrt{3}$? Can someone please elaborate. Many thanks.
 [/INST] As Shayan says,
$e^{i(-1 + i \sqrt{3})} = e^{-\sqrt{3} - i} = e^{-\sqrt{3}} e^{-i} = e^{-\sqrt{3}}(\cos(1) - i\sin(1)); \tag{1}$
it is also true that
$e^{i(-1 + i \sqrt{3})} = \cos(-1 + i \sqrt{3}) + i\sin(-1 + i \sqrt{3}), \tag{2}$
since for any $z \in \Bbb C$,
$e^{iz} = \cos z + i \sin z; \tag{3}$
in (3), $e^{iz}$, $\cos z$, and $\sin z$ are understood to be complex analytic (holomorphic) functions; (3) may be verified for $z \in \Bbb C$ via the power series expansions of these three entire functions about $0$:
$e^{iz} = \sum_0^\infty \dfrac{(iz)^n}{n!} = 1 + iz + i^2 \dfrac{z^2}{2} + i^3 \dfrac{z^3}{6} + \ldots = 1 + iz - \dfrac{z^2}{2} - i\dfrac{z^3}{6} + \ldots$
$= (1 - \dfrac{z^2}{2} + \ldots) + i(z - \dfrac{z^3}{6} + \ldots) = \cos z + i \sin z, \tag{4}$
etc. etc. etc.
</s>