Uniqueness of fractional linear transformations for ad-bc=1 Show that any fractional linear transformation can be represented in the form $$f(z) = \frac{az+b}{cz+d}$$
where $ad-bc = 1$.  Is this representation unique?
This is just the definition of a fractional linear transformation. I am having trouble proving that $ad-bc=1$.  I started from trying to take the derivative, but that got me nowhere.
I also have trouble understanding intuitively if the representation is unique or not.  Thank you!

If you replace the coefficieints $a,b,c,d$ by $ta,tb,tc,td$  where $t \neq 0$ you get the same function $f$. When you do this $ad-bc$ becomes $t(ad-bc)$. So  there is no question  of proving that if $f$ has the given form then $ad-bc$ must be $1$. What is true is that you can always take $t=\frac 1 {ad-bc}$ so that in the new form $ad-bc=1$, provided $ad-bc \neq 0$. Once you put the restriction that $ad-bc=1$ the coefficients become unique.
Suppose $\frac {az+b} {cz+d} =\frac {a'z+b'} {c'z+d'} $ for all $z$. Cross multiply and equate coefficients of $1,z$ and $z^{2}$. Armed with $ad-bc=1=a'd'-b'c'$ it is failry easy to see that $a=a',b=b',c=c'$ and $d=d'$.
