Is $e^{i+\pi}$ irrational or not? Since we know that the value of $e$, $i$, and $\pi$ are irrational reals, how about $$e^{i+\pi}\;?$$ Is it still irrational (that is, not a Gaussian rational)?
The problem make me curious until now.
 A: If you mean rational in the usual sense $-$ as a subset of the reals $-$ then $e^{i+\pi}$ is certainly not rational since it is not real. But the question of whether it's a Gaussian rational $-$ that is, its real and complex parts are rational $-$ may not be easy to prove. Notice that
$$e^{i+\pi} = e^{\pi}\cos 1 + i e^{\pi}\sin 1$$
Given irrational numbers $\alpha$ and $\beta$ it's often quite hard to determine whether $\alpha^{\beta}$ and $\alpha \beta$ (and $\alpha + \beta$) are rational or irrational. For example, it is not currently known whether $e\pi$ is rational or irrational.
A: Building off of Clive's answer, if $e^\pi \cos 1$ and $e^\pi \sin 1$ were both rational numbers, then so would be their quotient $\tan 1$.  But $\tan 1$ is irrational as a consequence of the Lindemann–Weierstrass theorem, because that theorem implies that $e^{2i}$ is transcendental, and because $e^{2i}=\dfrac{i-\tan 1}{\tan 1+i}$.  Therefore $e^{\pi+i}$ cannot have both its real and imaginary parts rational.  
A: $e^i=\cos1+i\sin1$. (By Euler's Formula)
So $e^{i+\pi}=e^\pi\cos1+ie^\pi\sin1$, which is non-real and, of course, irrational.
