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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.

Please share your thoughts. Thank you.

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  • $\begingroup$ 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). $\endgroup$
    – Jean Marie
    Apr 18 '20 at 11:28
  • $\begingroup$ What has this problem to do with Riemann surfaces? Or with Complex Analysis, for that matter? $\endgroup$ Apr 18 '20 at 11:29
  • $\begingroup$ @JoséCarlosSantos I have edited the question. Thank you. $\endgroup$ Apr 18 '20 at 11:31
  • $\begingroup$ @JeanMarie Thank you. Can you please explain to me, why it is unbounded? $\endgroup$ Apr 18 '20 at 11:42
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  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 $t\in\Bbb R$.
  3. You are right here.
  4. It is unbounded, since it contains every par $(\cosh t,\sinh t)$, with $t\in\Bbb R$.
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  • $\begingroup$ Thank you. Why $(z,\sqrt z)$ belongs to the set given in first option? $\endgroup$ Apr 18 '20 at 11:32
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    $\begingroup$ @JeanMarie Sure! I've edited my answer. Thank you. $\endgroup$ Apr 18 '20 at 11:50
  • $\begingroup$ 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 $d((0,0),(z,\sqrt{1-z^2}) = \sqrt{z^2 - \sqrt{1-z^2}}) = 1$. $\endgroup$ Apr 18 '20 at 12:37
  • $\begingroup$ I am getting $d((0,0),(z,\sqrt{1-z^2})) = \sqrt{z^2 - (\sqrt{1-z^2})^2} = 1$. $\endgroup$ Apr 18 '20 at 12:43
  • $\begingroup$ Which distance $d$ are you using in $\Bbb C^2$? $\endgroup$ Apr 18 '20 at 13:03

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