### 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 $L^{+}$.
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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.
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