$\newcommand{\C}{\mathbb{C}}\newcommand{\cn}{\colon}\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\sm}{\setminus}$I'm trying to compute the integral cohomology ring of $X=\{(x,y)\in\C^2\cn (x-iy)(x+iy)=x^2+y^2\ne 0\}=U\cap V$, where $$U=\{(x,y)\in\C^2\cn x\ne iy\}\sim\C^2\setminus\C\sim\R^4\setminus\R^2\sim S^{4-2-1}=S^1\sim V=\{(x,y)\in\C^2\cn x\ne -iy\},$$
since $\C^2\setminus U$ and $\C^2\setminus V$ are lines in $\C^2$. It is seen that $U\cup V=\C^2\sm \{(0,0)\}\sim\R^4\sm\{(0,0,0,0)\}\sim S^3$. Then from Mayer-Vietoris sequences one get $H_1(X)=\Z^2,$ $H_2(X)=\Z=H_0(X)$, since $X$ is path-connected, and $H_n(X)=0$ for $n>2$. Now, from duality, $H^n(X)=H_n(X)$ for all $n$. Let $a,b$ be generators of $H^1(X)$, and $c$ be generator of $H^2(X)$. It is evident that cup products $ac=bc=a^2=b^2=c^2=0$. But how one can determine $ab=kc$, where $k\in\Z$?
Note that the space $X$ is homotopy equivalent to the knot complement of a hopf link in $S^3$. Therefore, $X\simeq T^2=S^1\times S^1$, and the cohomology ring structure follows, i.e., $$H^*(X;\Bbb Z)\cong\Lambda_\Bbb Z[a,b]$$ where $a,b$ generate $H^1$. This tells us that $a\smile b$ generates $H^2$, so $a\smile b=\pm c$. (The sign depends on the choice of orientation.)
To see the homotopy equivalence, we consider a deformation retraction from $X$ to $S^3\setminus(M\cup N)$ by normalizing the magnitude of every vector (sliding each point $\vec{x}$ towards $S^3$ via the straight line connecting $\vec x$ and the origin), where $M=\{(x,y)\in\Bbb C^2\mid x/y=i\}\cap S^3$ and $N=\{(x,y)\in\Bbb C^2\mid x/y=-i\}\cap S^3$. Clearly, $M\cup N$ is a hopf link as they can be realized as the preimage $h^{-1}(\{i\}), h^{-1}(\{-i\})$ of the hopf map $h:S^3\to \Bbb C^*=\Bbb C\cup\{\infty\},$ $ (x,y)\mapsto x/y$.
To see that $X\simeq T^2$, we note that $S^3\setminus(M\cup N)$ is homeomorphic to $\Bbb R^3$ with a line and a circle around that line removed, which is homotopy equivalent to $T^2$. This is by considering the stereographic projection.