Graph on complex plane $x^2+y^2=1$

Question

Our professor told us that any affine plane curve in complex plane is unbounded so I am wondering what is the graph of $$x^2+y^2-1=0$$

Is it not a circle?

Answer 1

Let $\gamma = \{ (x,y) \in \mathbb{C}^2 | F(x,y)=0\}$, where $F \in \mathbb{C}[x,y]$ is a non-constant polynomial. Note that, for (almost) all $b \in \mathbb{C}$ the polynomial $F(\cdot, b) \in \mathbb{C}[x]$ is non-constant, so it has a complex solution $a$, since $\mathbb{C}$ is algebraically closed.

This means that for "arbitrarily large" $b\in \mathbb{C}$ (large in modulus) we can find some $(a,b) \in \gamma$. But this implies that $\gamma$ is unbounded.

For example, let $F=x^2+y^2 = (x+ i y)(x-iy)$. Then all points of the form $(in, n)$ belong to the curve defined by $F$ (because $(in)^2+n^2=0$).

In your example, $x^2+y^2-1 =0$ defines another curve $\Gamma$. For all $n \in \mathbb{N}$ you can solve the equation $x^2 = 1-n^2$ and find that the point $(i\sqrt{(n^2-1)}, n)$ belongs to $\Gamma$.