Question

Prove that the linear fractional transformation equation maps the circle equation into itself. Also prove

 

that f(z) is conformal in equation.

09 Jan 2026
Answer :
Word Count : 163
Consider the linear fractional transformation [ \phi(z)=\frac{2z-1}{2-z}. ] First, we show that the circle (C:\ |z|=1) is mapped into itself. Let (|z|=1). Then (\bar z = 1/z). Compute [ |\phi(z)|^2=\phi(z)\overline{\phi(z)} =\frac{2z-1}{2-z}\cdot \frac{2\bar z-1}{2-\bar z}. ] Substituting (\bar z=1/z), [ |\phi(z)|^2 =\frac{(2z-1)\left(\frac{2}{z}-1\right)}{(2-z)\left(2-\frac{1}{z}\right)}. ]
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