# Bounded subsets of $\Bbb C^2$.

Which of the following subsets of $$\Bbb C^2$$ are bounded?

1. $$\{(z,w): z^2 + w^2 = 1\}$$,

2. $$\{(z,w): |Re z|^2 + |Re w|^2 = 1\}$$,

3. $$\{(z,w): |z|^2 + |w|^2 = 1\}$$,

4. $$\{(z,w): |z|^2 - |w|^2 = 1\}$$.

For option $$1$$, I found some elements in the set but all have norm $$1$$. But still, I am not convinced.

option $$3$$ is the unit ball in $$\Bbb R^4$$ hence it is bounded. Other options I am not sure.

• The first set is unbounded : in particular, you cannot say the norm is 1 (don't forget that the nome is a $\ge 0$ number). Apr 18 '20 at 11:28
• What has this problem to do with Riemann surfaces? Or with Complex Analysis, for that matter? Apr 18 '20 at 11:29
• @JoséCarlosSantos I have edited the question. Thank you. Apr 18 '20 at 11:31
• @JeanMarie Thank you. Can you please explain to me, why it is unbounded? Apr 18 '20 at 11:42

1. It is unbounded, since it contains every pair of the form $$(z,w)$$, where $$w$$ is a square root of $$1-z^2$$.
2. It is unbounded, since it contains every pair $$(1,it)$$, with $$t\in\Bbb R$$.
4. It is unbounded, since it contains every par $$(\cosh t,\sinh t)$$, with $$t\in\Bbb R$$.
• Thank you. Why $(z,\sqrt z)$ belongs to the set given in first option? Apr 18 '20 at 11:32
• Thank you. But is it immediate that set of all $(z,\sqrt{1-z^2})$ is unbounded? For example, in real case, such tuples are bounded. I am getting $d((0,0),(z,\sqrt{1-z^2}) = \sqrt{z^2 - \sqrt{1-z^2}}) = 1$. Apr 18 '20 at 12:37
• I am getting $d((0,0),(z,\sqrt{1-z^2})) = \sqrt{z^2 - (\sqrt{1-z^2})^2} = 1$. Apr 18 '20 at 12:43
• Which distance $d$ are you using in $\Bbb C^2$? Apr 18 '20 at 13:03