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).
By the Conformal Mapping Theorem, f(-i) = 2 + i.
Well i'm not sure what that is, and also, 2+i seems to be wrong.
3/5+4/5i
+157 
