Show that $ad-bc \ne 0$ for a composition of two Mobius transformations.

Question

I could prove that a composition of two Mobius transformations is again a Mobius transformation. Let be $T(z)=\dfrac{a_1z+b_1}{c_1z+d_1}, \ (a_1d_1-c_1b_1 \ne 0)$ and $S(z)=\dfrac{a_2z+b_2}{c_2z+d_2}, \ (a_2d_2-c_2b_2 \ne 0)$, so $S[T(z)]=\dfrac{a_3z+b_3}{c_3z+d_3}$. But I can't prove whether $a_3d_3-c_3b_3 \ne 0)$. The book says $a_3d_3-c_3b_3 = (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ but in order that to hold also must $c_2b_2(c_1b_1+a_1d_1)=0$ hold to make $a_3d_3-c_3b_3 - (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ to be zero. I did calculations thrice! Where am I doing wrong?

Answer 1

Note that\begin{align}S\bigl(T(z)\bigr)&=\frac{a_2\frac{a_1z+b_1}{c_1z+d_1}+b_2}{c_2\frac{a_1z+b_1}{c_1z+d_1}+d_2}\\&=\frac{a_2(a_1z+b_1)+b_2(c_1z+d_1)}{c_2(a_1z+b_1)+d_2(c_1z+d_1)}\\&=\frac{(a_2a_1+b_2c_1)z+a_2b_1+b_2d_1}{(c_2a_1+d_2c_1)z+c_2b_1+d_2b_1}\end{align}

So,\begin{align}\begin{vmatrix}a_3&b_3\\c_3&d_3\end{vmatrix}&=\begin{vmatrix}a_2a_1+b_2c_1&a_2b_1+b_2d_1\\c_2a_1+d_2c_1&c_2b_1+d_2b_1\end{vmatrix}\\&=\det\left(\begin{bmatrix}a_2&b_2\\c_2&d_2\end{bmatrix}.\begin{bmatrix}a_1&b_1\\c_1&d_1\end{bmatrix}\right)\\&=\begin{vmatrix}a_2&b_2\\c_2&d_2\end{vmatrix}.\begin{vmatrix}a_1&b_1\\c_1&d_1\end{vmatrix}\\&\neq0.\end{align}

Answer 2

There is a natural identification of these transformations with that of $2\times 2$ matrices with non-zero determinant:

$$\frac{az+b}{cz+d}\longleftrightarrow \begin{pmatrix}a&b\\ c&d \end{pmatrix}.$$

You can, relatively easily, prove that composition of two of these transformations coincides with matrix multiplication, which makes the result easier to see.