Is $\{ (x, y) \in \mathbb{C}^2 : x^2 + y^2 = 1\}$ connected in $\mathbb{C}^2$ ?
How do I approach this problem ?
I think that that the fact that continuous functions map connected sets to connected sets might be helpful.
But I am having trouble using it. What function should I use ?
I thought of $f:\mathbb{C} \to \mathbb{C}^2$ defined as $f(z) = (z, \sqrt{1-z^2})$. But that won't suffice since there are points of the form $(z, -\sqrt{1-z^2})$ as well. What should I do ?
Also, what can we say about the path-connectedness of the set ?
Any help shall be highly appreciated.
You can use the stereographic projection which holds for any field $K$ (your case being that of $K=\mathbb{C}$).
For $(x,y)\in K^2$ with $x^2+y^2=1$ and $y\not=1$, setting $t=\frac{x}{1-y}$, one gets $$x=\frac{2t}{t^2+1},\ y=\frac{t^2-1}{t^2+1}$$
Geometric insights can be found there (well, it is 3-dimensional but you can put $z=0$ for your 2-dimensional case). Then any point of your complex curve $x^2+y^2=1$ with $y\not=1$ can be connected to $(0,-1)$ ($t=0$), $(1,0)$ ($t=1$) and $(-1,0)$ ($t=-1$). If your starting point is $(0,1)$ then exchange $x$ and $y$ and use connection with $(0,-1)$ as intermediate point. Then, through this very intuitive parametrization, you get the fact that your curve (the complexified cercle) is connected.