Complex Variables and Partial Differential Equations (3130005)

Question-2a

BE | Semester-3 Winter-2019 | 26-11-2019

Q2) (a)

Determine the Mobius transformation which maps $z_{1} = 0, z_{2} = 1, z_{3} = \infty$ into $, w_{1} = - 1, w_{2} = - i, w_{3} = 1$ .

z_{1} = 0, z_{2} = 1, z_{3} = \infty, w_{1} = - 1, w_{2} = - i, w_{3} = 1 .

The cross ratio,

\frac{(w - w_{1}) (w_{2} - w_{3})}{(w - w_{3}) (w_{2} - w_{1})} = \frac{(z - z_{1}) (z_{2} - z_{3})}{(z - z_{3}) (z_{2} - z_{1})}

\Rightarrow \frac{(w + 1) (- i - 1)}{(w - 1) (- i + 1)} = \lim_{z_{3} \to \infty} \frac{(z - 0) (1 - z_{3})}{(z - z_{3}) (1 - 0)}

\Rightarrow - \frac{(w + 1) (1 + i)}{(w - 1) (1 - i)} = \lim_{z_{3} \to \infty} \frac{z z_{3} (\frac{1}{z_{3}} - 1)}{z_{3} (\frac{1}{z_{3}} - 1)}

\Rightarrow - \frac{(w + 1)}{(w - 1)} i = \lim_{z_{3} \to \infty} \frac{z (0 - 1)}{(0 - 1)} (∵ \frac{(1 + i)}{(1 - i)} = i & z_{3} \to \infty \Rightarrow \frac{1}{z_{3}} \to 0)

\Rightarrow \frac{(w + 1)}{(w - 1)} = - \frac{z}{i}

\Rightarrow \frac{(w + 1)}{(w - 1)} = zi (∵ \frac{1}{i} = - i)

\Rightarrow \frac{(w + 1) + (w - 1)}{(w + 1) - (w - 1)} = \frac{zi + 1}{zi - 1} (∵ Componendo and dividendo)

\Rightarrow \frac{2 w}{2} = \frac{zi + 1}{zi - 1}

\Rightarrow w = \frac{zi + 1}{zi - 1}

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