Question

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?

• A. \(S_1\) is true.
• B. \(S_2\) is true.
• C. \(S_2\) is true and \(S_1\) is false.
• D. Neither \(S_1\) is true nor \(S_2\) is true.

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.

Answer 1

The Correct Option is B, C

Step 1: Understanding the Question.
We need to find the actual image regions \(f(D)\) and \(g(D)\) under the two given Mobius maps, then compare them to check the two inclusions \(S_1\) and \(S_2\). A Mobius transformation always sends a disc or a half-plane to another disc or half-plane, so we look for which half-plane or disc each map produces.

Step 2: Find f(D).
Let \(w = f(z) = \dfrac{z}{1-z}\). Solve for \(z\) in terms of \(w\):
\[ w(1-z) = z \implies w = z + wz = z(1+w) \implies z = \dfrac{w}{1+w} \]
The condition \(|z| < 1\) becomes \(\left| \dfrac{w}{1+w} \right| < 1\), that is \(|w| < |w+1|\). This says \(w\) is closer to \(0\) than to \(-1\). The set equidistant from \(0\) and \(-1\) is the vertical line \(\text{Re}(w) = -\tfrac12\), and the side closer to \(0\) is \(\text{Re}(w) > -\tfrac12\). So
\[ f(D) = \left\{ w : \text{Re}(w) > -\tfrac12 \right\} \]

Step 3: Find g(D).
Let \(w = g(z) = \dfrac{1+z}{1-z}\). Solve for \(z\):
\[ w(1-z) = 1+z \implies w - 1 = z + wz = z(1+w) \implies z = \dfrac{w-1}{w+1} \]
The condition \(|z|<1\) becomes \(|w-1| < |w+1|\), meaning \(w\) is closer to \(1\) than to \(-1\). The set equidistant from \(1\) and \(-1\) is the imaginary axis \(\text{Re}(w) = 0\), and the side closer to \(1\) is \(\text{Re}(w) > 0\). So
\[ g(D) = \left\{ w : \text{Re}(w) > 0 \right\} \]

Step 4: Compare the two half-planes.
\(f(D)\) is the half-plane \(\text{Re}(w) > -\tfrac12\), and \(g(D)\) is the smaller half-plane \(\text{Re}(w) > 0\). Every point with \(\text{Re}(w) > 0\) also satisfies \(\text{Re}(w) > -\tfrac12\), so
\[ g(D) \subseteq f(D) \]
This means \(S_2\) is TRUE.

Step 5: Check S1.
Is \(f(D) \subseteq g(D)\)? Take \(w = -\tfrac14\), which satisfies \(\text{Re}(w) = -\tfrac14 > -\tfrac12\), so it lies in \(f(D)\). But \(\text{Re}(w) = -\tfrac14\) is not greater than \(0\), so this point is NOT in \(g(D)\). This single point already shows \(f(D)\) is not contained in \(g(D)\), so \(S_1\) is FALSE.

Step 6: Match to the options.
Since \(S_2\) is true, option (B), "\(S_2\) is true," is TRUE. Since \(S_2\) is true and \(S_1\) is false, option (C) is also TRUE (it is a more detailed, equally correct restatement). Option (A), "\(S_1\) is true," is FALSE, and option (D), "neither is true," is FALSE since \(S_2\) does hold.

Final Answer:
\(g(D) \subseteq f(D)\) but \(f(D) \not\subseteq g(D)\), so \(S_2\) is true and \(S_1\) is false.
\[ \boxed{\text{(B) and (C)}} \]