Is $e^{2\pi i m/k}$ a $k$th root of unity even if $\gcd (m,k)>1$? At the beginning of Ch.8 of the book Apostol's Analytic Number Theory, it is written:

[consider] ... the exponential function $$f(n) = e^{2\pi i mn/k} $$ 
  where $m$ and $k$ are fixed integers. The number $e^{2\pi i m/k}$ is a $k$th root of unity and $f(n)$ is its $n$th power. 

Is this true if $\gcd (m,k)>1$? Isn't it necessary for $m$ to be either prime or $1$? 
For example let $k=2m$ and $m$ is a prime, then the number $e^{2\pi i m/k}$ will be $\frac{k}2$th of root of unity. Later on the book it appears that the $\gcd (m,k)=1$ is never considered. 
 A: 2 is the square root of 4, it is also the fourth root of 16. 1 is the cube root of 1, it is also the twelfth root of 1. All that matter is that if you multiply the number by itself $k$ times, you get the number it's meant to be the $k$th root of.
A: In the situation  you're worried about, where the gcd of $k$ and $m$ is some $d>1$, you'd have $m=dm'$ and $k=dk'$ for some $m'$ and $k'$ whose gcd is 1.  Since $m/k=m'/k'$, your $e^{2\pi inm/k}$ equals $e^{2\pi inm'/k'}$, where your worries have vanished because the gcd of $m'$ and $k'$ is $1$.
A: The $k$-th roots of unity are k, as already explained in other answer.
Those which are not also the roots of a lower $j < k$, i.e. when $\gcd(m,k)=1$ with your symbology, are called Primitive k-th Roots of 1.
A: All that is necessary (and sufficient) for $z$ to be a $k$th root of unity is for $z^k = 1$. That is all.
$$(e^{2 \pi i m / k})^k = e^{2 \pi i m} = 1$$
So they're all $k$th roots. 
If, for example, $k = a \cdot m$, then it will also be true that it's an $a$th root:
$$(e^{2 \pi i m / k})^a = (e^{2 \pi i m/(a\cdot m)})^a = 1$$
So given $k = q \cdot m$, then for which $b \in \mathbb N$ is this a $b$th root?
$$(e^{2 \pi i m / k})^b = (e^{2 \pi i /q})^b = 1 \quad\text{ iff }\quad b/q \in \mathbb Z \; .$$
Now $b/q = b/(k/m) = bm/k$.
This tells us that $e^{2 \pi i m /k}$ is a $b$th root precisely when $b$ is a multiple of $k/m$. So what is the smallest integral multiple of $k/m$? It is $k/\mathrm{gcd}(k,m)$. To see this, first imagine simplifying the fraction by eliminating common factors: this is $\mathrm{gcd}(k,m)$. Then we multiply by the denominator, which is now $m/\mathrm{gcd}(k,m)$. We multiply this by $k/m$ to obtain its smallest integral multiple.
In conclusion, $e^{2 \pi i m / k}$ is a $b$th root for precisely the following $b$:
$$b \in \{ n \cdot k / \mathrm{gcd}(k,m) : n \in \mathbb N\}$$
