= Solution
The difference map $(z,w)\mapsto z-w$ is <continuous> and $X$ is its <inverse element>[inverse] image of the <open set> $\mathbb C\setminus\{0\}$, so $X$ is open in $\mathbb C^2$. For fixed $z$, the <equivalence class> is $\{z\}\times(\mathbb C\setminus\{z\})$, <homeomorphic> to a punctured plane. It is <path-connected>: two points can be joined by a polygonal path avoiding the missing point. Thus all <equivalence classes> are <connected>.
The projection $\pi$ is <continuous> and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in $\mathbb C^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi(U)$. Hence $\pi(U)$ is open.
Its fibers are precisely the <equivalence classes>, so it induces a <bijection> $\bar\pi:X/R\to\mathbb C$. <Continuity> follows from the <quotient topology> and $\pi=\bar\pi\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar\pi(V)=\pi(q^{-1}(V))$ is open. Thus the <inverse element>[inverse] is <continuous> too, proving $\boxed{X/R\cong\mathbb C}$ as <topological spaces>. The plane is <connected>, so the <connected>-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.
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