Exercise · ZW-MOBIUS-TRANSFORMATION-008 · ★★☆ Intermediate
Which Möbius transformation sends \(0\mapsto1\), \(1\mapsto0\) and \(\infty\mapsto\infty\)?
Choose one option.
Show answer
- D. \(1-z\)
Explanation
Since \(f(\infty)=\infty\), we need \(c=0\) and \(f(z)=az+b\). From \(f(0)=1\) we get \(b=1\), and from \(f(1)=0\) we get \(a=-1\).
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-005 The inverse of f(z)=(2z+1)/(z+1) is f^-1(z)=
- 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-009 What does the circle |z|=1 become under f(z)=(z+1)/(z-1)?
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