Homotopy Equivalence between $S^1$ and the space of quadratic polynomials. Homotopy Equivalence between $S^1$ and the space of quadratic polynomials $z^2 + pz + q$ with complex $p$ and $q$ with distinct roots.
I have tried $p^2 - 4q > 0$, since the fundamental theorem of Algebra guarantee one solution. 
This is a upper half of a parabola in $q-p$ plane, how is it homotopy equivalence to $S^1$?

Let $X$ be the space of monic quadratic polynomials with distinct roots. The easiest way to show that sending a polynomial to $p^2-4q$ creates a homotopy equivalence is to think about it geometrically. In particular, note that the space of monic quadratic polynomials with distinct roots is equivalent to the space of unordered distinct pairs in $\mathbb C$, since such a polynomial is described uniquely by its roots.
Looking at the quadratic formula, if we let $r_1$ and $r_2$ be the roots of a polynomial $z^2+pz+q$, we note that $p^2-4q$ is equal to $\frac{1}4(r_1-r_2)^2$. Note that, from the givens, you know precisely that $r_1-r_2$ is not zero, but it could be any other complex number - so $p^2-4q$ could, similarly, be any non-zero complex number.
So, the expression $p^2-4q$ defines a continuous function $X\rightarrow \mathbb C\setminus \{0\}$. You can, however, figure out that this is actually a homotopy equivalence: first, one can map $\mathbb C\setminus \{0\}$ back into $X$ by a function $g$ that takes $\alpha\in \mathbb C\setminus\{0\}$ to $x^2-\frac{1}4\alpha$. Note that $f\circ g$ is the identity on $\mathbb C\setminus\{0\}$.
Then, we just need to show that $g\circ f$ is homotopic to the identity on $X$ - and we can do that by noting that, if $P\in X$ then the roots $r_1$ and $r_2$ of $P$ and the roots $r'_1$ and $r'_2$ of $g\circ f(P)$ share the quantity $(r_1-r_2)^2 = (r'_1-r'_2)^2$, which means that $r_1-r_2 = \pm (r'_1-r'_2)$ - in particular, that the roots are in a parallelogram. To form a homotopy from $g\circ f$ to the identity, one simply needs to interpolate between these unordered pairs $(r_1,r_2)$ and $(r'_1,r'_2)$ uniformly over all polynomials - but this is easy since they are in a parallelogram and we can simply act by translating the pair $(r'_1,r'_2)$ over time towards $(r_1,r_2)$.
Then, to finish, you just note that defining $h : \mathbb C \setminus \{0\} \rightarrow S^1$ to be the function that takes $z$ to $\frac{z}{\|z\|}$ gives a homotopy equivalence of those two spaces, which can be composed with the one you already found.
