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 \(f(-i)\).
Thanks!
