De Moivre's Theorem Related - Complex number According to de Moivre's Theorem: If $n$ is any positive integer then: $(\cos\theta + i\sin\theta)^n=\cos n\theta +i\sin n\theta$
Also $(\cos\theta +i\sin \theta)^{\frac{1}{n}} = \cos \frac{2r\pi +\theta}{n}+i\sin \frac{2r\pi+\theta}{n}$ where $r = 0,1,2,\ldots, n-1.$
Please suggest why the last part which is: $\cos \frac{2r\pi +\theta}{n}+i\sin \frac{2r\pi+\theta}{n}$ is not only $(\cos \frac{1}{n} \theta +i\sin\frac{1}{n}\theta)$?
 A: Using Euler Formula, $e^{ix}=\cos x+i\sin x$
if $(e^{i\theta})^\frac1n=e^{i\alpha},$  
So, $e^{i\theta}=e^{in\alpha}\implies e^{i(n\alpha-\theta)}=1=e^{2ir\pi}$ where $r$ is any integer
So, $n\alpha-\theta=2r\pi\implies \alpha=\frac{2r\pi+\theta}n$
Now, if $e^{i\alpha_1}=e^{i\alpha_2},\implies e^{i(\alpha_1-\alpha_2)}=1=e^{2is\pi}$ where $s$ is any integer
$\implies \alpha_1-\alpha_2=2s\pi \implies \frac{2r_1\pi+\theta}n-\frac{2r_2\pi+\theta}n=2s\pi\implies r_1-r_2=s\cdot n$
So, we have $n$ in-congruent values of $r\pmod n$ corresponding to $n$ in-congruent roots of $(e^{i\theta})^\frac1n$ as all the roots are distinct.
A: $z^{\frac{1}{n}}$ is defined as a set of points $w$ such that $w^n = z$. Apply the first version of the formula for $z^n$ to the formula for $z^{\frac{1}{n}}$ and you will see why the second formula works.
A: For $z\in\Bbb C$ the terms $z^5$, or $e^z$, are well defined, but $z^{1/n}$ is not. Writing $z^{1/n}$ is shorthand for "I'm interested in the set $\{w\in\Bbb  C\ |\ w^n=z\}$". Depending on the problem you are working on you might need all of these $w$, or only a specific one that satisfies some other conditions present in your situation.
A: A possible way to look at the situation is this:
The first version of de Moivre's theorem could be written as:
$$(\cos\theta + i\sin\theta)^n=\cos (2r\pi+n\theta) +i\sin (2r\pi+n\theta)$$
but there wouldn't be much point, as all the values of $r$ you could choose would all give the same answer, since $\sin$ and $\cos$ both have period $2\pi$.
On the other hand, in the version:
$$(\cos\theta +i\sin \theta)^{\frac{1}{n}} = \cos \frac{2r\pi +\theta}{n}+i\sin \frac{2r\pi+\theta}{n}$$
choosing different values for $r$ do not all give the same values for the right-hand side, and in fact choosing $n$ different values for $r$ will give you the $n$ possible $n$-th roots of $\cos\theta + i\sin\theta$ ... so long as you make sure the values you choose are all non-congruent $\pmod{n}$.
A: the last part are the $n$th-roots of the number $e^{i \theta}$. If you power all of those numbers to the $n$th-power you will get again $e^{i \theta}$.
A: If you solve for cos(t)= 1/2 you get not only t= + / - pi/3 as  solutions, but all coterminal angles of vector in the complex Argand diagram get included as a general solution for periodic functions. 
