Connectedness of $\mathbb{C}^{n}\setminus \mathbb{R}^{n}$ Hi everyone: How would you prove this classic result: if $ n\geq2 ,$  then $\mathbb{C}^{n}\setminus \mathbb{R}^{n}$ is connected? Any reference? Thanks.

The set that you are interested in is path connected.  Consider $(z_1,\cdots,z_n),(w_1,\cdots,w_n)\in\mathbb{C}^n\setminus\mathbb{R}^n$.  Since $(z_1,\cdots,z_n)\in\mathbb{C}^n\setminus\mathbb{R}^n$ there is at least one index $i$ where $z_i$ is not real (some of the other indices can be real, points in $\mathbb{R}^n$ have all coordinates real).  Similarly, there is some index $j$ where $w_j$ is not real.  We construct two paths.  There are two cases:
Case 1: $i\not=j$.  
Step 1: While keeping $z_i$ fixed, transform all $z_1,\cdots,z_{i-1},z_{i+1},\cdots,z_n$ into $w_1,\cdots,w_{i-1},w_{i+1},\cdots,w_n$.  This can be done with a linear homotopy $(1-t)z_k+tw_k$ changes $z_k$ into $w_k$ as $t$ varies between $0$ and $1$.  This never intersects $\mathbb{R}^n$ since the $i$-th coordinate is never real.
Step 2: While keeping all other coordinates fixed, transform $z_i$ into $w_i$.  Since the $j$-th coordinate is not real, this path never intersects $\mathbb{R}^n$.
Case 2: $i=j$.  In this case, let $k$ be an index other than $i$.  Consider the path that keeps all other coordinates fixed and transforms $z_k$ into the complex number $0+1i$ ($i$ is not an index here).  Then use Case 1 on $k$ and $j$.
