NBHM 2020 PART C Question 24 Solution ($B = \{(z,w) \in \Bbb C^2 : |Re\, z|^2 + |Re\, w|^2 = 1\}$ is not bounded )

Listed below are four subsets of  $\Bbb C^2$. For each of them, write “Bounded” or “Unbounded” in the box as the case may be. ($Re(z)$ denotes the real part of a complex variable $z$).
1. $A = \{(z,w) \in \Bbb C^2 : z^2 + w^2 = 1\}$.
2. $B = \{(z,w) \in \Bbb C^2 : |Re\, z|^2 + |Re\, w|^2 = 1\}$.
3. $C = \{(z,w) \in \Bbb C^2 : |z|^2 + |w|^2 = 1\}$.
4. $D = \{(z,w) \in \Bbb C^2 : |z|^2 - |w|^2 = 1\}$.
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Solution: Consider the standard metric in $\Bbb C$ given by $d(z,w) = |z-w|$. Now, the metric on $\Bbb C^2$ is given by $$d((z_1,w_1),(z_2,w_2)) = \sqrt{|z_2 - z_1|^2 + |w_2 - w_1|^2}.$$ 
Option 1:(Unbounded) We have $\{(it, \sqrt{1-(it)^2}) : t \in \Bbb R\} \subseteq A$. Now, $$d((0,0),(it,\sqrt{1+t^2})) = \sqrt{(t^2)+(1+t^2)} \ge \sqrt{t^2} =  t.$$ and can be made arbitrarily large. So $A$ is unbounded.
Option 2:(Unbounded) $\{(1, it) : t \in \Bbb R\} \subseteq B$. Now, $$d((0,0),(1, it)) = \sqrt{(1^2)+(t^2)} = \sqrt{1+t^2} \ge t.$$ which can be made arbitrarily large. So $B$ is unbounded.
Option 3:(Bounded) $C = \{(z,w) \in \Bbb C^2 : |z|^2 + |w|^2 = 1\} = \{(z,w) : d((0,0),(z,w)) = 1\}$. This is the unit sphere in $\Bbb C^2$ and hence $C$ is bounded.
Option 4:(Unbounded) We have $cosh^2\,t - sinh^2\,t = 1$ for $t \in \Bbb R$. Hence $\{(cosh\,t, sinh\,t) : t \in \Bbb R\} \subseteq D$. Now, $$d((0,0),(cosh\,t, sinh\,t)) = \sqrt{(cosh^2\,t)+(sinh^2\,t)} \\ = \sqrt{(\frac{e^t + e^{-t}}{2})^2+(\frac{e^t-e^{-t}}{2})^2} = \sqrt{\frac{e^{2t}+e^{-2t}}{2}}.$$ which can be made arbitrarily large. So $D$ is unbounded.
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