Where's the flaw in this application of De Moivre's fomula to find n-th roots? To find the $n^\text{th}$ roots of a complex number, we can first express it in polar form (I'm assuming $r=1$ for brevity; it doesn't matter for my question):
\begin{align}
e^{i\theta} &= \cos\theta + i\sin\theta \\
             &= \cos\left(\theta + 2\pi k\right) + i\sin\left(\theta + 2\pi k\right), \qquad k\in\mathbb{Z}
\end{align}
and then use De Moivre's formula:
\begin{align}
\left(e^{i\theta}\right)^\frac{1}{n} &= \cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right), \qquad k\in\mathbb{Z}
\end{align}
However, when we express a complex number in polar form we could also shift the $\cos$ and $\sin$ arguments by different multiples of $2\pi$; this is certainly still valid:
\begin{align}
e^{i\theta} &= \cos\theta + i\sin\theta \\
             &= \cos\left(\theta + 2\pi a\right) + i\sin\left(\theta + 2\pi b\right), \qquad a,b\in\mathbb{Z}
\end{align}
but it does it not follow that:
\begin{align}
\left(e^{i\theta}\right)^\frac{1}{n} &= \cos\left(\frac{\theta + 2\pi a}{n}\right) + i\sin\left(\frac{\theta + 2\pi b}{n}\right), \qquad a,b \in\mathbb{Z}
\end{align}
This is obviously false; for example, taking $\theta = 0$, $n=3$, and setting $a=1$, $b=0$, we get $-0.5 + 0i$ as a cube root of unity.
I know that the answer has to be "this is not a correct application of De Moivre's formula", but an explanation of why this is the case would be nice.
 A: Since $z=\cos\frac{\theta+2\pi}3 + i\sin\frac{\theta}3$ is no longer a valid polar form, this $z$ cannot be considered as an answer of the following question:

Find $z$ such that $z^n =\cos\theta + i\sin\theta$.

Because de Moivre's formula only applies to polar forms:
$$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$$
A: 
and then use De Moivre's formula: \begin{align}
\left(e^{i\theta}\right)^\frac{1}{n} &= \cos\left(\frac{\theta + 2\pi
k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right), \qquad
k\in\mathbb{Z} \tag1\end{align}


this is certainly still valid:
\begin{align}
e^{i\theta}   &= \cos\left(\theta + 2\pi a\right) + i\sin\left(\theta + 2\pi b\right), \qquad a,b\in\mathbb{Z}\tag2\end{align}
but it does it not follow that: \begin{align}
\left(e^{i\theta}\right)^\frac{1}{n} &= \cos\left(\frac{\theta + 2\pi
a}{n}\right) + i\sin\left(\frac{\theta + 2\pi b}{n}\right), \qquad a,b\in\mathbb{Z}\tag3 \end{align}

Neither $(1)$ nor $(3)$ is an application of De Moivre's theorem, which does not generally hold for non-integer powers.
$(1)$ expresses the multi-valued definition of $e^z$, and can be proven for real $\theta$ and nonero $n$ ($n$ can even be complex).
On the other hand, there is no particular reason for $(3)$ to follow from $(2);$ $(3)$ is indeed not generally true, as evidenced by your counterexample.
