### CSIR JUNE 2011 PART C QUESTION 81 SOLUTION ($f(z) = \frac{z}{3z+1}$ maps $H^+$ to $H^{+}$)

Define $H^{+} = \{z=x+iy \in \Bbb C : y > 0\}$
$H^{-} = \{z=x+iy \in \Bbb C : y < 0\}$
$L^{+} = \{z=x+iy \in \Bbb C : x > 0\}$
$L^{-} = \{z=x+iy \in \Bbb C : x < 0\}$
Then the bilinear transformation $$f(z) = \frac{z}{3z+1}$$ maps
1. $H^{+}$ onto $H^{+}$ and $H^{-}$ onto $H^{-}$.
1. $H^{+}$ onto $H^{-}$ and $H^{-}$ onto $H^{+}$.
1. $H^{+}$ onto $L^{+}$ and $H^{-}$ onto $L^{-}$.
1. $H^{+}$ onto $L^{-}$ and $H^{-}$ onto $L^{+}$.
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:
Option 1: (True) We have $$f(z) = \frac{z}{3z+1} = (3+\frac{1}{z})^{-1}.$$
Let $z=x+iy \in H^{+}$ then $\frac{1}{z} = \frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2}$ whose imaginary part is negative. Hence $\frac{1}{z}(H^{+}) = H^{-}$. The function $g(z) = z + 3$ preserves all four regions since it is just a translation. Therefore if $z \in H^{+}$ then $\frac{1}{z} \in H^{-}$ and $\frac{1}{z}+3 \in H^{-}$. This shows that, $f(z) = (\frac{1}{z}+3)^{-1} \in H^{+}$. Therefore $f$ sends $H^+$
onto $H^{+}$. Similarly if $z \in H^{-}$ then $\frac{1}{z} \in H^{+}$ and $\frac{1}{z}+3 \in H^{+}$. This shows that, $f(z) = (\frac{1}{z}+3)^{-1} \in H^{-}$. Therefore $f$ sends $H^{-}$ onto $H^{-}$.

Option 2:(False)
Option 3:(False)
Option 4:(False)
We have proved that $f$ sends $H^+$ onto $H^{+}$ and $H^{-}$ onto $H^{-}$. Hence they cannot be mapped to any other sets. Hence all the remaining options are false.
### 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...