If $\theta\in\mathbb{Q}$, is it true that $(\cos \theta + i \sin \theta)^\alpha = \cos(\alpha\theta) + i \sin(\alpha\theta)$? Is the following true if $\theta\in\mathbb{Q}$?
$$(\cos \theta + i \sin \theta)^\alpha = \cos(\alpha\theta) + i \sin(\alpha\theta)$$
Is it true if $\alpha\in\mathbb{R}$? In each case, prove or give a counterexample, whichever is applicable. 
I am not able to guess except about the de Moivre's theorem.
 A: It is not true in general.
$$
(\cos \theta + i \sin \theta)^\frac 12 = \left [ \begin{array}{l}
\cos \frac \theta 2 + i \sin \frac \theta 2 \\
\cos \left ( \frac \theta 2 + \pi \right ) + i \sin \left ( \frac \theta 2 + \pi \right )
\end{array}\right .
$$
A: The problem with this question is that this is not so much about being true as about being well defined (except in the cases where $\alpha\in\Bbb Z$, for which there is no problem and the formula holds).
Raising complex numbers to non-integral powers is not well defined; the best stab at a definition is $z^\alpha=\exp(\alpha\ln z)$ which shows that the necessary branch cut of the complex logarithm makes it impossible to define $z\mapsto z^\alpha$ as a (single-valued) continuous function on all of$~\Bbb C$, or even on a neighbourhood of$~0$, unless $\alpha\in\Bbb Z$. Therefore the left hand side $\def\i{\mathbf i}(\cos θ + \i \sin \theta)^\alpha$ of your proposed equality has these kind of definitional problems, whereas the right hand side $\cos(\alpha\theta) + \i\sin(\alpha\theta)=\exp(\i\alpha\theta)$ is a perfectly well defined (even for $\theta\in\Bbb C$ of which it defines a holomorphic function).
As you can easily check you can make the formula hold for a given $\theta$ by choosing $\ln(\cos θ + \i\sin \theta)=\i\theta$ in defining the LHS, which then becomes $\exp(\alpha\mathbf i\theta)$ in accordance with the RHS. However, to get the formula to still work after adding $2\pi$ to $\theta$ would require using a different logarithm for the same complex number $\cos θ + \i\sin\theta$; no single-valued choice for defining complex exponentiation can have it both ways. In a sense the position that makes the equation $(\cos\theta + \i\sin\theta)^\alpha = \cos(\alpha\theta) + \i\sin(\alpha\theta)$ "as true as possible" for $\alpha\notin\Bbb Z$, is to interpret the left hand side as a multi-valued expression (finitely many if $\alpha\in\Bbb Q$, infinitely many otherwise), one of whose values coincides with the single-values right hand side. However working with multi-valued expressions is not what one does normally in mathematics, it is therefore a tricky business, and I think it tends to obscure rather than to solve fundamental difficulties, so one is better off not doing this.
A: Not for every $\alpha \in \mathbb{R}$, but for the left hand side of your equation you can calculate
\begin{eqnarray}
(\cos(\theta) + i \sin(\theta))^\alpha & = & (e^{i\theta})^\alpha = (e^{i\theta + 2\pi i k})^\alpha = e^{(i\theta + 2\pi i k)\alpha} = e^{i\theta\alpha + 2\pi i k \alpha} \\
& = & e^{i\theta \alpha} e^{2\pi i k \alpha} = e^{2\pi ik \alpha} e^{i \theta \alpha} = e^{i \pi 2 k \alpha} e^{i \alpha \theta} \\
& = & e^{i \pi 2k \alpha} (\cos(\alpha \theta) + i \sin(\alpha\theta))
\end{eqnarray}
for $k \in \mathbb{Z}$ so the left hand side is not necessarily unique, but at least one of its values has the value of the right hand side. It is found by choosing $k = 0$. A counterexample is found by choosing $k = 1$ and $\alpha = \frac{1}{2}$ as in Kaster's answer.
