### 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. $D = \{(z,w) \in \Bbb C^2 : |z|^2 - |w|^2 = 1\}$.
I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.
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.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...