For which $n\in\mathbf{N}$ do we have $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$? Put $z_{n} = e^{2\pi i /n}$. I am searching for $n \in \mathbf{N}$ so that $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$.
I know that : $z_{5} = \cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5}) $ and $z_{7} =\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})$.
Can you give me a hint how to continue my search? Thank you. 

That's one too many hints in the comments, but the OP still seems in doubt, so here is a proof that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_{35})$, where $\zeta_n=e^{2\pi i/n}$ is a primitive $n$th root of unity.
First, let us show that $\mathbb{Q}(\zeta_5,\zeta_7)\subseteq\mathbb{Q}(\zeta_{35})$. Notice that 
$$\zeta_{35}^7=(e^{2\pi i/35})^7 = e^{2\pi i/5}=\zeta_5.$$
Thus, $\zeta_5\in \mathbb{Q}(\zeta_{35})$. Similarly, $\zeta_7 = \zeta_{35}^5 \in \mathbb{Q}(\zeta_{35})$. Hence, $\mathbb{Q}(\zeta_5,\zeta_7)\subseteq\mathbb{Q}(\zeta_{35})$.
Next, let us show that $\mathbb{Q}(\zeta_{35})\subseteq \mathbb{Q}(\zeta_5,\zeta_7)$. Indeed, consider 
$$(\zeta_5\cdot\zeta_7)^3 = (e^{2\pi i/5}\cdot e^{2\pi i/7})^3 = (e^{2\pi i\cdot 12/35})^3 = e^{2\pi i\cdot 36/35} = e^{2\pi i}\cdot e^{2\pi i/35} = 1 \cdot e^{2\pi i/35} = \zeta_{35}.$$
Thus, $\zeta_{35}=(\zeta_5\cdot\zeta_7)^3\in \mathbb{Q}(\zeta_5,\zeta_7)$, and this shows the inclusion $\mathbb{Q}(\zeta_{35})\subseteq \mathbb{Q}(\zeta_5,\zeta_7)$. Therefore, we must have an equality of fields.
Now, suppose that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_n)$ for some $n\geq 1$. We have just shown that $n=35$ works. Are there any other possible values of $n$ that work? Well, if $n$ is odd, then $\mathbb{Q}(\zeta_n) = \mathbb{Q}(\zeta_{2n})$, so $n=70$ also works. 
Finally, one can show that if $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$, then $m$ divides $n$ (here neither $m$ or $n$ should be twice an odd number). In particular, since we know that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_{35})=\mathbb{Q}(\zeta_n)$, then $n$ is divisible by $35$, and therefore $\varphi(n)$ is divisible by $24$. If $n>70$ and divisible by $35$, then $\varphi(n)$ would be strictly larger than $24$, and that would be a contradiction, because $\varphi(n)$ is the degree of the extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$. Hence, $n=35$ or $70$.
