TITLE: Is the graph of $xy=1$ in $\mathbb C^{2}$ connected?
QUESTION [5 upvotes]: The graph  of  $xy=1$  in  $\mathbb C^{2}$ is  set  of  points $(x+iy,u+iv)$ that satisfies  $$xu-yv=1$$ and $$uy+xv=0$$ How  to  find  if  this  set  is  connected  or  not . 
I  also  have another  doubt. Can  this  set  mean  the collection  of  points $\{x+iy \in \mathbb C | xy=1\}$=$\{x+{{i}\over {x}}|x\in \mathbb R\}$ $?$  
Some  lead  please ,  I  am  totally  clueless.

REPLY [2 votes]: It is easiest to show that this is path connected.
Step 1: Prove that every point in the solution set is connected to a point where $|x|=1$.  Namely, if $xy=1$, then consider multiples $x/\lambda$ and $\lambda y$, but varying $\lambda$ between $1$ and $|x|$, you can continuously change $xy=1$ into a solution where $|x|=1$.
Step 2: Observe that when $|x|=1$, $|y|=\frac{1}{|x|}=1$ as well.  Therefore, it is enough to prove that the solution set where $|x|=1$ is also path connected.  For every $x=e^{i\theta}$ it must be that $y=e^{-i\theta}$ (which is the only solution).  By varying $\theta$, you construct a path from any point $x$ on the unit circle to any other point.