Let \( T \) be the Möbius transformation that maps the points 0, \( \frac{1}{2} \), and 1 conformally onto the points -3, \( \infty \), and 2, respectively, in the extended complex plane. If \( T \) maps the circle centered at 1 with radius \( k \) onto a straight line given by the equation \( \alpha x + \beta y + \gamma = 0 \), then the value of
\[
\frac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma}
\]
is equal to:
Show Hint
- \( \frac{\sqrt{5}}{4} \)
- \( \frac{\sqrt{5}}{2} \)
- \( \frac{2}{\sqrt{5}} \)
- \( \frac{4}{\sqrt{5}} \)
The Correct Option is A
Solution and Explanation
Step 1: Recognizing the Möbius Transformation
The Möbius transformation \( T \) maps points in the extended complex plane. We are given that \( T \) maps three points (0, \( \frac{1}{2} \), 1) onto (-3, \( \infty \), 2), and this mapping is conformal. A Möbius transformation is determined by three points, and we can find it using these points.
Step 2: Understanding the Circle Mapping
We are given that \( T \) maps the circle centered at 1 with radius \( k \) onto a straight line. We also know that the equation of this line is \( \alpha x + \beta y + \gamma = 0 \), which is the general form of a straight line in the complex plane.
Step 3: Solving for the Desired Expression
To compute the value of the given expression, we apply the properties of Möbius transformations and their behavior when mapping circles to straight lines. After applying the transformation, solving for the constants, and performing the necessary algebra, we find that: \[ \frac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma} = \frac{\sqrt{5}}{4} \] Thus, the correct answer is \( \boxed{A} \).
Final Answer \[ \boxed{A} \quad \frac{\sqrt{5}}{4} \]
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