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I'm trying to get the hang of this theorem by solving some exercises involving it. One of those says: Prove that $(\mathbb{Q},+)/\mathbb{Z}$ is isomorphic to $(U_\infty,\cdot)$, where $U_\infty = \{ z \in C\mid z^n = 1, n \in \mathbb{N} \}$.

I know I need to find a morphism from $(\mathbb{Q},+)$ to $(U_\infty,\cdot)$ such that $\ker(f) = \mathbb{Z}$. I believe that $f(x)=\cos(2\pi x)+i \sin(2\pi x)$ does the trick, but I'm not sure. Could anyone tell me if this is correct, and if not, give another example of a function?

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    You're on the right way. Now show that $f$ is a group morphism, that $\ker f=\mathbb Z$ and that $f$ is onto. Now, what does the First Isomorphism theorem states? – 2017-01-29
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    I have already shown that $f$ is a group morphism, that $Ker(f)=Z$. The first isomorphism theorem states that if there is a morphism between two groups G and K, then there exists an isomorphism f from $G/Ker(f)$ to $Im(f)$. – 2017-01-29
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    Now, prove that $Im (f)=U_\infty$. Do you need any help with it? – 2017-01-29
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    If you could, that would be great – 2017-01-29

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I'm going to provide a full answer to the problem. The morphism $f:(\mathbb Q,\rightarrow (U_\infty, \cdot), f(x)=(\cos(2\pi x), \sin(2\pi x))$ is a group morphism and $\ker f=\mathbb Z$.

Now let us see that $f$ is onto. Choose $z\in U_\infty$, and we want to find $x\in\mathbb Q$ so that $f(x)=z$. As $z\in U_\infty$, there exists $n\in\mathbb N$ so that $z^n=1$. Now, let us write $z=\cos (\theta) + i\sin(\theta)$. As $z^n=1$, $|z|^n=1$ so $|z|=1$. We have

$$z^n=1=\cos(n\theta)+i\sin(n\theta) \Rightarrow \cos n\theta =1 $$ so $n\theta= 2\pi m$ with $m\in\mathbb N$, and $\theta=\frac{2m}{n}\pi$.

That means $$z=\cos\left(\frac{2m}{n}\pi\right), \sin\left(\frac{2m}{n}\pi\right) = f\left(\frac mn\right)$$ as we wanted to prove.