Monomorphism and Epimorphism Example 6.9, pg 241, Number Theory and Abstract Algebra, Cracking the GRE Math Subject Test (4th Edition):

Let $U$ be the multiplicative group of the $n$th roots of unity; this group is cyclic of order $n$ and is generated by $w=\cos\left(\frac{2\pi}{n}\right)+i\sin\left(\frac{2\pi}{n}\right)$. If we define $\Phi:(\Bbb Z,+)\to U$ by the equation $\Phi(a)=w^a$, show that $\Phi$ is a homomorphism. Is $\Phi$ a monomorphism? An epimorphism? An isomorphism?

QUESTION
I know $\Phi$ is homomorphic. I also know that $\Phi$ cannot be isomorphic because $\Bbb Z$ is infinite while $U$ is finite and an infinite group cannot be isomorphic to a finite one. However, the textbook claims $\Phi$ is not monomorphic (and I think this is correct) by using the concept of the triviality of the kernel of $\Phi$ but in this case, I do not really understand how they used that to arrive at their conclusion. We know that if $\Phi(a)=\Phi(b)$ implies $a=b$, then $\Phi$ is monomorphic. This appears to be true in this case, a contradiction to what the textbook claims (again, I do not claim to be correct). So, what I'm I missing here?
Lastly, the book claims $\Phi$ is epimorphic but does not show how or why. We know that if we can show that $\Phi(w^a)=a$, then the veracity of the claim is established. Again, I do not know how to prove this.
I will really appreciate it if someone can shed light on this for me. I will also welcome different approaches that can be used to solve these problems apart from the ones discussed above.
 A: Re 

We know that if $\Phi(a)=\Phi(b)$ implies $a=b$, then $\Phi$ is monomorphic. This appears to be true in this case, a contradiction to what the textbook claims.

No, this simply is not true: $\Phi(0)=\Phi(n)$, by (what in an exp-free-context is often called) de Moivre's formula, yet $0\neq n$ in $\mathbb{Z}$, hence it is indeed not a monomorphism. 
EDIT: re your request for "different approaches": this problem that this textbook is posing to you is so narrow that there hardly is any leeway for "different approaches", which is why you will not get many, I think. The best I can think of is to tell you that $w = \cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n}) = \exp(i\frac{2\pi}{n})$, where $\exp$ denotes the exponential function. It is a better "method" to treat such problems using $\exp$, already for the psychological reason that this way, you will have to work with one argument-parenthesis-pair instead two. (Note also the redundancy in having to write $\frac{2\pi}{n}$ twice when working with the raw de Moivre's formula only.) Using $\exp$ can justifiably be called a "method" for example because it allows you to use the laws for exponentiation to e.g. see why $\Phi(n)=\Phi(0)$, since $\Phi(n)=w^n=(\exp(i\frac{2\pi}{n}))^n=(\mathrm{e}^{i\frac{2\pi}{n}})^n=\mathrm{e}^{i\frac{2\pi}{n}\cdot n} = \mathrm{e}^{i\cdot2\pi}$ $=$ (periodicity of $\exp$) $=$ $\mathrm{e}^{i\cdot 0}$ $=$ $\mathrm{e}^{0}$ $=$ $1$ $=$ $\Phi(0)$. If you want, you can view what you just witnessed to be a toy example of a method, in the sense that it took me no conscious thinking to turn the crank (as they say) and grind out the preceding calculation. In principle your request for "different methods" is commendable, this problem however happens to be too easy for such a request be appropriate. I hope this helps. Best wishes.
