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To elaborate on Diego Matessi's answer (and Boyarsky's comment which I only saw after writing this)... the map $F$ is symmetric, so it factors as a composite map $$C^2\longrightarrow (C^2)_{\Sigma_2}\longrightarrow C^2.$$ Here$\Sigma_2$ is the group with $2$ elements acting on $C^2$ by switching coordinates, the first map is the quotient map, and the second map is well-known to be a homeomorphism. It can be thought of as the map that sends an unordered pair $[z_1, z_2]$ to the unique monic complex polynomial whose roots are $-z_1$ and $-z_2$. So the map $S^3\to F(S^3)$ is not injective. Nevertheless, it is an easy exercise to show that $F(S^3)$ is homeomorphic to $S^3$. But then again, it {\em seems} to me that $F(S^3)$ should not be a smooth submanifold of ${\mathbb C}^2$. $F(S^3)$ can be identified with the space of monic complex quadratic polynomials $z^2+bz+c$ whose pair of roots gives a unit complex vector. So it is the space of pairs of complex numbers $(b, c)$ that solve the equation $$|\frac{-b+\mbox{first square root of } b^2-4c}{2}|^2+|\frac{-b+\mbox{second square root of } b^2-4c}{2}|^2=1.$$ It seems that the space of solutions of this equation should not be smooth, and that a singularity should occur where the discriminant $b^2-4c$ is zero - which corresponds to the diagonal in the original $C^2$, but I could be mistaken.

Edit Let me point out that the equation can be simplified. Let $\Delta=b^2-4c$. Then the equation is in fact equivalent to $$(*) \,\, |b|^2+|\Delta|=2.$$ Since the map $(b,c)\to (b, b^2-4c)$ is a diffeomorphism, the question of whether $F(M)$ is a smooth submanifold is equivalent to the question whether the space of solutions of the equation ($\ast$) is a smooth submanifold of $C^2$. I just wanted to point this out because equation ($\ast$) looks simpler than the one I originally wrote. In the meantime, Sergei Ivanov gave an argument showing that it is not smooth.

3 added 140 characters in body

To elaborate on Diego Matessi's answer (and Boyarsky's comment which I only saw after writing this)... the map $F$ is symmetric, so it factors as a composite map $$C^2\longrightarrow (C^2)_{\Sigma_2}\longrightarrow C^2.$$ Here$\Sigma_2$ is the group with $2$ elements acting on $C^2$ by switching coordinates, the first map is the quotient map, and the second map is well-known to be a homeomorphism. It can be thought of as the map that sends an unordered pair $[z_1, z_2]$ to the unique monic complex polynomial whose roots are $-z_1$ and $-z_2$. So the map $S^3\to F(S^3)$ is not injective. Nevertheless, it is an easy exercise to show that $F(S^3)$ is homeomorphic to $S^3$. But then again, it {\em seems} to me that $F(S^3)$ should not be a smooth submanifold of ${\mathbb C}^2$. $F(S^3)$ can be identified with the space of monic complex quadratic polynomials $z^2+bz+c$ whose pair of roots gives a unit complex vector. So it is the space of pairs of complex numbers $(b, c)$ that solve the equation $$|\frac{-b+\mbox{first square root of } b^2-4c}{2}|^2+|\frac{-b+\mbox{second square root of } b^2-4c}{2}|^2=1.$$ It seems that the space of solutions of this equation should not be smooth, and that a singularity should occur where the discriminant $b^2-4c$ is zero - which corresponds to the diagonal in the original $C^2$, but I could be mistaken.

To elaborate on Diego Matessi's answer (and Boyarski's Boyarsky's comment which I only saw after writing this)... the map $F$ is symmetric, so it factors as a composite map $$C^2\longrightarrow (C^2)_{\Sigma_2}\longrightarrow C^2.$$ Here$\Sigma_2$ is the group with $2$ elements acting on $C^2$ by switching coordinates, the first map is the quotient map, and the second map is well-known to be a homeomorphism. It is can be thought of as the map that sends an unordered pair $[z_1, z_2]$ to the unique monic complex polynomial whose roots are $-z_1$ and $-z_2$. So the map $S^3\to F(S^3)$ is not injective. Nevertheless, it is an easy exercise to show that $F(S^3)$ is homeomorphic to $S^3$. But then again, it {\em seems} to me that $F(3)$ F(S^3)$should not be a smooth submanifold of${\mathbb C}^2$.$F(S^3)$can be identified with the space of monic complex quadratic polynomials$z^2+bz+c$whose pair of roots gives a unit complex vector. So it is the space of pairs of complex numbers$(b, c)\$ that solve the equation $$|\frac{-b+\mbox{first square root of } b^2-4c}{2}|^2+|\frac{-b+\mbox{second square root of } b^2-4c}{2}|^2=1.$$ It seems that the space of solutions of this equation should not be smooth, but I could be mistaken.