IGNOU MMT 5 2026 - Prove that the linear fractional transformation maps the ci

Question

Prove that the linear fractional transformation $\phi(z) = \frac{2z - 1}{2 - z}$ maps the circle $c : |z| = 1$ into itself. Also prove that f(z) is conformal in $\bar{D} = \{z : |z| \leq 1\}$ .

10 Jan 2026

Answer :

Word Count : 141

Let (\phi(z)=\dfrac{2z-1}{2-z}). For (z) on the circle (|z|=1), consider [ |\phi(z)|^2=\frac{|2z-1|^2}{|2-z|^2}. ] Now [ |2z-1|^2=(2z-1)(2\bar z-1)=4z\bar z-2z-2\bar z+1, ] and [ |2-z|^2=(2-z)(2-\bar z)=4-2z-2\bar z+z\bar z. ] Since (|z|=1), we have (z\bar z=1). Substituting, [ |2z-1|^2=4(1)-2z-2\bar z+1=5-2z-2\bar z, ] [ |2-z|^2=4-2z-2\bar z+1=5-2z-2\bar z. ] Hence (|2z-1|^2=|2-z|^2), and therefore (|\phi(z)|^2=1).

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