Let $X_{n}$ be a set $$ X_{n} = \{(x_{1}, \dots, x_{n})\in \mathbb{C}^{n}\,:\, x_{1}^{2}+ \cdots + x_{n}^{2} = 1\}. $$ For $n\geq 2$. Then $X_{n}$ is connected.
In the case of $\mathbb{R}$, it is just a sphere and one can show that it is path connected. However, over $\mathbb{C}$, I don't know how to prove it in an easy way. It is known that if $f(X_{1}, \dots, X_{n})\in \mathbb{C}[X_{1}, \dots, X_{n}]$ is an irreducible polynomial over $\mathbb{C}$, then its zero set in $\mathbb{C}^{n}$ is connected with respect to usual topology on $\mathbb{C}^{n}$, which is not easy to prove. Since $X_{n}$ is just a set defined by deg 2 equation, there might be easy way to prove this.
Geometrically, $X_n(\mathbb{C})\subset\mathbb{R}^{2n}$ is the intersection of $\sum x_i^2-\sum y_i^2=1$ with $\sum x_iy_i=0$. This is an $S^{n-2}$-bundle over $y\in\mathbb{R}^n-\{0\}$ glued to an $S^{n-1}$ over the point $y=0$, so you can move around to the limit point $y=0$ and use the connectedness of the fiber $S^{n-1}$ over the origin to get $X_n$ connected.