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Recall that a Mobius transformation \(f\) has an equation of the form \(f(z)=\dfrac{az+b}{cz+d}\) where a, b, c, and d are complex numbers.

Suppose that \(f\) is a Mobius transformation such that \(f(1)=i\)\(f(i)=-1\), and \(f(-1)=1\). Find the value of \(f(-i)\).

 

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 Jan 31, 2021
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f(-i) = 1/2.

 Jan 31, 2021

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