c) Prove that the linear fractional transformation maps the circle c : | z | =1 into itself. Also prove that f (z) is conformal in D ={z:| z| 1}.

To prove that the linear fractional transformation Φ(z) = (2z-1)/(2-z) maps the circle c: |z| = 1 into itself, we'll show that all the points on the circle c are mapped back onto the circle c.

Let z = e^(iθ) be a point on the circle c, where θ is a real number. Then |z| = 1.

Now, we need to find Φ(z) and show that |Φ(z)| = 1.

Φ(z) = (2z - 1) / (2 - z) Φ(e^(iθ)) = (2e^(iθ) - 1) / (2 - e^(iθ))

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