For f(z)=(3z+1)/(3z-2) (z≠frac23), is it true that…

For $f(z)=\dfrac{3z+1}{3z-2}$ ($z\neq\frac23$), is it true that $f[f(z)]\equiv z$?

Step 1 · Compute $f[f(z)]$ directly

Let $w=f(z)=\dfrac{3z+1}{3z-2}$. Then:

$$ f(w)=\frac{3w+1}{3w-2} $$

Numerator: $\ 3w+1=\dfrac{3(3z+1)+(3z-2)}{3z-2}=\dfrac{12z+1}{3z-2}$. Denominator: $\ 3w-2=\dfrac{3(3z+1)-2(3z-2)}{3z-2}=\dfrac{3z+7}{3z-2}$.

$$ f[f(z)]=\frac{12z+1}{3z+7} $$

Step 2 · This is not $z$

$\dfrac{12z+1}{3z+7}=z$ would require $12z+1=z(3z+7)=3z^2+7z$, i.e. $3z^2-5z-1=0$ — true only for the two specific roots of this quadratic, not for all $z$. So $f[f(z)]\not\equiv z$.

Step 3 · Why: the matrix criterion for self-inverse Möbius maps

Writing $f(z)=\dfrac{az+b}{cz+d}$ as the matrix $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, composition of Möbius maps corresponds to matrix multiplication. $f$ is self-inverse ($f\circ f=\text{id}$) exactly when $M^2$ is a scalar multiple of the identity — which for a $2\times2$ matrix happens exactly when $\text{tr}(M)=a+d=0$ (by the Cayley–Hamilton theorem, $M^2=(a+d)M-(ad-bc)I$, so $M^2\propto I$ iff $a+d=0$).

Step 4 · Check the given $f$

Here $a=3,\ b=1,\ c=3,\ d=-2$, so $\ a+d=3+(-2)=1\neq0$.

$$ \boxed{f[f(z)]\not\equiv z \ \text{ for this } f\text{, since } a+d=1\neq0} $$

The problem statement's claim does not hold for the given $f$ — this appears to be an error in the original problem (perhaps a coefficient typo), rather than a computational mistake in the solution.

Step 5 · What a genuinely self-inverse map looks like

Any $f(z)=\dfrac{az+b}{cz-a}$ (note the $-a$, giving trace $a+(-a)=0$) is self-inverse. The simplest example, $f(z)=-\dfrac1z$ (i.e. $a=0,b=-1,c=1,d=0$, trace $0$):

$$ f[f(z)]=f\left(-\frac1z\right)=-\frac1{-1/z}=z \quad\checkmark $$

Verification

Numeric check of Step 1's formula at $z=1$: $f(1)=\dfrac{3+1}{3-2}=4$; $f(4)=\dfrac{12+1}{12-2}=\dfrac{13}{10}$. Direct formula: $\dfrac{12(1)+1}{3(1)+7}=\dfrac{13}{10}$ ✓ — matches, and $\frac{13}{10}\neq1$, confirming $f[f(1)]\neq1$.