Discussion
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 centred at 1 with radius k onto a straight line given by the equation \(\alpha x + \beta y + \gamma = 0\), then the value of \(\dfrac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma}\) is equal to
Answer: A
The value is \(\dfrac{1}{7}\).
Since \(\tfrac{1}{2} \mapsto \infty\), write \(T(z) = (az + b)/(2z - 1)\). Then \(T(0) = -b = -3\) gives b = 3, and \(T(1) = a + 3 = 2\) gives a = -1, so \(T(z) = (3 - z)/(2z - 1)\).
A circle maps to a line exactly when it passes through the point sent to \(\infty\), here 1/2; the circle centred at 1 must therefore have radius \(k = \tfrac{1}{2}\). It also passes through 3/2 and \(1 + \tfrac{i}{2}\), whose images are
\[ \begin{aligned} T(\tfrac{3}{2}) &= \tfrac{3}{4} \\ T(1 + \tfrac{i}{2}) &= \tfrac{3}{4} - \tfrac{5}{4}i \end{aligned} \]
so the image is the line \(x = 3/4\): \(\alpha = 1\), \(\beta = 0\), \(\gamma = -3/4\). Then
\[ \begin{aligned} \frac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma} &= \frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} \\ &= \frac{1}{7} \end{aligned} \]
The ratio does not depend on how the line's equation is scaled.
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