Let \(f(z) = \dfrac{z}{1-z}\) and \(g(z) = \dfrac{1+z}{1-z}\) be two Mobius transformations defined on the unit disc \(D = \{z \in \mathbb{C} : |z| < 1\}\). Consider the following statements:
\(S_1\): \(f(D) \subseteq g(D)\)
\(S_2\): \(g(D) \subseteq f(D)\)
Which of the following statements is/are CORRECT?
Show Hint
Both maps send \(z=1\) to infinity, so the unit circle maps to a vertical line in each case; find that line for \(f\) and \(g\) using \(z=0\) and \(z=-1\), then compare the two half-planes.
Step 1: Understanding the Concept.
A Mobius map sends circles and lines to circles and lines, and it sends the boundary of a disc (here the unit circle $|z|=1$) either to a circle or to a straight line in the $w$-plane. Once we know where the boundary and one interior point go, we know exactly which side of the boundary the whole disc maps to.
Step 2: Locate the image of the disc under f.
For $f(z) = \dfrac{z}{1-z}$, the only point of the extended plane sent to infinity is $z=1$, which lies ON the unit circle, so the boundary circle $|z|=1$ must map to a straight line under $f$. Test the interior point $z=0$: $f(0) = 0$. Test a boundary point, $z=-1$: $f(-1) = \dfrac{-1}{1-(-1)} = \dfrac{-1}{2} = -\tfrac12$. Since $f(0)=0$ is an interior image and $f(-1)=-\tfrac12$ is a boundary image, and the boundary must be a vertical line by symmetry of the map about the real axis, that line is $\text{Re}(w) = -\tfrac12$, with the interior point $0$ on the side $\text{Re}(w) > -\tfrac12$. So $f(D) = \{\text{Re}(w) > -\tfrac12\}$.
Step 3: Locate the image of the disc under g.
For $g(z) = \dfrac{1+z}{1-z}$, again $z=1$ is the point sent to infinity, so the boundary again maps to a line. Test $z=0$: $g(0) = \dfrac{1+0}{1-0} = 1$. Test $z=-1$: $g(-1) = \dfrac{1-1}{1+1} = 0$. So the boundary line passes through $w=0$ (image of $z=-1$), and by the same symmetry it must be the imaginary axis $\text{Re}(w)=0$, with the interior image $g(0)=1$ on the side $\text{Re}(w) > 0$. So $g(D) = \{\text{Re}(w) > 0\}$.
Step 4: Compare the two half-planes.
$g(D) = \{\text{Re}(w)>0\}$ sits entirely inside $f(D) = \{\text{Re}(w) > -\tfrac12\}$, since $\text{Re}(w)>0$ automatically gives $\text{Re}(w)>-\tfrac12$. So $g(D) \subseteq f(D)$, meaning $S_2$ is true.
Step 5: Rule out S1.
The strip $-\tfrac12 < \text{Re}(w) \le 0$ belongs to $f(D)$ but not to $g(D)$, so $f(D)$ is strictly bigger than $g(D)$ and cannot be contained in it. So $S_1$ is false.
Step 6: Final Answer.
Since $S_2$ is true, option (B) holds; and since $S_2$ is true while $S_1$ is false, option (C) also holds. Options (A) and (D) both fail.
\[ \boxed{\text{(B) and (C)}} \]
