Question

Let \( T \) be a Möbius transformation such that \( T(0) = \alpha, T(\alpha) = 0 \) and \( T(\infty) = -\alpha \), where \( \alpha = \frac{-1 + i}{\sqrt{2}} \). Let \( L \) denote the straight line passing through the origin with slope \( -1 \), and let \( C \) denote the circle of unit radius centered at the origin. Then, which of the following statements are TRUE?

• A. \( T \) maps \( L \) to a straight line
• B. \( T \) maps \( L \) to a circle
• C. \( T^{-1} \) maps \( C \) to a straight line
• D. \( T^{-1} \) maps \( C \) to a circle

Hint:

In Möbius transformations, straight lines map to straight lines, and circles map to circles. The inverse transformation reverses this mapping.

Answer 1

The Correct Option is A, C

We are given a Möbius transformation \( T \) defined by the following conditions: \[ T(0) = \alpha, \quad T(\alpha) = 0, \quad T(\infty) = -\alpha, \] where \( \alpha = \frac{-1 + i}{\sqrt{2}} \). We also have a straight line \( L \) with slope \( -1 \) passing through the origin, and the circle \( C \) with unit radius centered at the origin. Option (A) \( T \) maps \( L \) to a straight line: This statement is true. A Möbius transformation generally maps lines and circles to other lines and circles. Since \( L \) is a straight line, \( T \) will map \( L \) to another straight line. The mapping behavior depends on the specific parameters of the transformation, but the general property of Möbius transformations ensures that lines map to lines in this case. Thus, option (A) is correct. Option (B) \( T \) maps \( L \) to a circle: This statement is false. As explained earlier, Möbius transformations map straight lines to straight lines or circles to circles, depending on the specific transformation. Since \( L \) is a straight line, \( T \) will map it to another straight line, not a circle. Thus, option (B) is incorrect. Option (C) \( T^{-1} \) maps \( C \) to a straight line: This statement is true. The inverse of a Möbius transformation \( T^{-1} \) also maps lines and circles to other lines or circles. Since \( T \) maps the circle \( C \) to a circle (as Möbius transformations map circles to circles), the inverse \( T^{-1} \) will map the circle to a straight line. Thus, option (C) is correct. Option (D) \( T^{-1} \) maps \( C \) to a circle: This statement is false. As explained, the inverse \( T^{-1} \) will map the circle \( C \) to a straight line, not a circle. Thus, option (D) is incorrect. Therefore, the correct answers are (A) and (C).