Suppose that $f$ is a Mobius transformation such that $f(1)=i$, $f(i)=-1$, and $f(-1)=1$. Find the value of $f(-i)$.
It seems like a simple problem, but I have great difficulty getting anywhere at all. A few pointers, and perhaps a complete solution, would be of much help to me; thanks!
EDIT: Sorry for the late edit! At the request of saulspatz, the equation for a Mobius transformation is $\frac{az+b}{cz+d}$. And yes, I am learning about Mobius transformations in a precalculus class.
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Your three data points give three equations, but the form of the Mobius transformation has four unknowns. The trick is that, unless $f$ happens to be linear, $c$ will not be zero. That means we can divide the top and bottom of the fraction by $c$:
$$f(z) = \frac{az + b}{cz + d} = \frac{\frac{a}{c}z + \frac{b}{c}}{z + \frac{d}{c}}$$
This form only has three unknowns, currently called $\frac{a}{c}$, $\frac{b}{c}$, and $\frac{d}{c}$. But since all of the unknowns are allowed to take any value, we might as well think of them as $a$, $b$, and $d$ respectively -- in other words, assuming that $c = 1$. The conclusion is that if $f$ is not linear, you may assume $c = 1$.
If $f$ were linear, two points would be enough to pin it down; but I'll leave it to you to verify that the linear transformation given by the first two points doesn't match the third.
Now you have three equations and three unknowns, which provides a system of equations you can solve.
A Mobius transformation is of the form $$ f(z)=\frac{az+b}{cz+d}\;. $$ Imposing the conditions you get $$ i=f(1)=\frac{a+b}{c+d}\\ -1=f(i)=\frac{ai+b}{ci+d}\\ 1=f(-1)=\frac{b-a}{d-c} $$ which is equivalent to a system of 3 equations in 4 unknowns; but since you will put these numbers in the linear fractional map, everything will be fine.