Question

Prove that the linear fractional transformation \Phi (z) = \frac{2z-1}{2-z} maps the circle c : \lvert z \rvert =1 into itself. Also prove that f(z) is conformal in \overline{D} = \left \{ z : \lvert z \rvert \leq 1 \right \}.

07 Feb 2024
Answer :
Word Count : 466

To prove that the linear fractional transformation \(\Phi(z) = \frac{2z-1}{2-z}\) maps the unit circle \(C: |z| = 1\) into itself, we need to show that if \(|z| = 1\), then \(|\Phi(z)| = 1\).

Let \(z = e^{i\theta}\) be a point on the unit circle, where \(\theta\) is a real parameter.

Then, 
\[
\Phi(z) = \frac{2e^{i\theta} - 1}{2 - e^{i\theta}} = \frac{2e^{i\theta} - 1}{2 - e^{i\theta}} \cdot \frac{2 - e^{-i\theta}}{2 - e^{-i\theta}} = \frac{(2e^{i\theta} - 1)(2 ______ ___ ________ _______ __________ ___ _____ __________ _____ ________ ____ ________.
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