Exercise · ZW-MOBIUS-TRANSFORMATION-005 · ★☆☆ Basic
The inverse of \(f(z)=\dfrac{2z+1}{z+1}\) is \(f^{-1}(z)=\)
Choose one option.
Show answer
- B. \(\dfrac{z-1}{2-z}\)
Explanation
From \(w=\frac{2z+1}{z+1}\): \(w(z+1)=2z+1\) gives \(z(w-2)=1-w\), that is \(z=\frac{w-1}{2-w}\). This amounts to inverting the coefficient matrix.
Review the concept: Möbius transformation
A Möbius transformation is a function of the extended complex plane to itself given by a quotient of two linear expressions with complex coefficients \(a,b,c,d\), arranged so that it is not constant.
\[f(z)=\frac{az+b}{cz+d},\qquad ad-bc\neq0\]
More practice problems: möbius transformation
- ZW-MOBIUS-TRANSFORMATION-001 Evaluate f(z)=(z+1)/(z-1) at z=i.
- ZW-MOBIUS-TRANSFORMATION-002 Which of these functions is NOT a Möbius transformation?
- ZW-MOBIUS-TRANSFORMATION-003 If f(z)=(2z+1)/(z+3), what is f(∞)?
- ZW-MOBIUS-TRANSFORMATION-004 Which point does f(z)=(z+2)/(3z-6) send to ∞?
- ZW-MOBIUS-TRANSFORMATION-006 If f(z)=1z and g(z)=z+1, what is g(f(z))?
- ZW-MOBIUS-TRANSFORMATION-007 Find the fixed points of f(z)=(z+2)/(z+1).
- ZW-MOBIUS-TRANSFORMATION-008 Which Möbius transformation sends 0mapsto1, 1mapsto0 and ∞mapsto∞?
- ZW-MOBIUS-TRANSFORMATION-009 What does the circle |z|=1 become under f(z)=(z+1)/(z-1)?
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