Check whether given set $\left \{ (z,w) \in \mathbb{C}^2 : z^2+w^2=1\right \}$ is bounded or unbounded

Question

The given set is $$\left \{ (z,w) \in \mathbb{C}^2 : z^2+w^2=1 \right \}$$

I have to check that is this set is bounded or unbounded?

solution i tried-here the given condition is $$z^2+w^2=1$$

i am considering $z=x+iy$ ,and $w=a+ib$ so from this we get
$$\Rightarrow (x+iy)^2+(a+ib)^2=1$$ $$\Rightarrow x^2-y^2+2xyi+a^2-b^2+2abi=1$$

equating real and imaginary part

$$ \Rightarrow x^2-y^2+a^2-b^2=1,\;\;2xy+2ab=0$$

$$\Rightarrow x^2-y^2+a^2-b^2=1,\;\;xy+ab=0$$

$$x^2-y^2+a^2-b^2=1,\;\;xy=-ab$$

after this i tried but does't getting any idea how to proceed further

Please Help

Thank you

Answer 1

Define $z=a-bi$ and $w=a+bi$, hence $$z^2+w^2=1\implies a^2-b^2={1\over2}$$which is a hyperbola w.r.t. $a$ and $b$ over which they can grow unboundedly and consequently $|z-w|=2|b|$. Therefore, the set is unbounded.

Answer 2

For every $z\in \mathbb C$, you always have some $w=\sqrt{1-z^2}\in\mathbb C$.
As $z$ is unbounded so is the pair $(z,w)\in\mathbb C^2$.

Answer 3

Here is a definitive answer to all similar questions:
If a subset $X\subset \mathbb C^n \; (n\geq 1)$ is given by the simultaneous vanishing of a family $(f_i)_{i\in I}$ of holomorphic functions $f_i:\mathbb C^n \to \mathbb C$, i.e. $X=\{z\in \mathbb C^n \vert \forall i\in I, f_i(z)=0\}$, then we have the alternative: $$\mathbf { X\operatorname {is either finite or else is unbounded}}$$ This is because a Stein variety can only be compact if it is finite.