Since, $e^{i\pi}=-1$,
right part of the equation ($-1$) is real and left part is seemingly complex as
$e^{ix}=\cos(x)+i\sin(x)$ which is a complex number( I am not sure though).
I am a freshman, and don't know much regarding Euler's equations and advanced Mathematics.
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$\begingroup$Real numbers are also complex numbers; their imaginary part is $0$. If $x$ is a real number then $e^{ix}$ is, in general, complex, but, in special cases, it can be real. The only real numbers of the form $e^{ix}$ ($x$ real) are $1$ and $-1$.
$\endgroup$ $\begingroup$Because $e^{ix}=\cos(x) + i\sin(x)$, it is only "real" when $\sin(x) = 0$, namely when $x = n \pi$ for $n \in \mathbb{Z}$. For every other value of $x$, $e^{ix}$ has a complex component. Note that I put "real" with quotes because $\mathbb{R} \subset \mathbb{C}$.
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