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A Mobius Transformation f has a 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).

 Jan 21, 2021
 #1
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By the Conformal Mapping Theorem, f(-i) = 2 + i.

 Jan 22, 2021
 #2
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Well i'm not sure what that is, and also, 2+i seems to be wrong.

tanmai79  Jan 22, 2021
 #3
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3/5+4/5i

 Jan 23, 2021

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