Answer
The function has a discontinuity at $ z=-2$ and $ z=3$. Moreover, the discontinuity is infinite because $\lim\limits_{x \to 2}\frac{1-2z}{z^2-z-6}=-\infty $ .
Work Step by Step
Since $$ z^2-z-6=0 \Longrightarrow (z+2)(z-3)=0\Longrightarrow z=3, z==-2$$
then the function has a discontinuity at $ z=-2$ and $ z=3$. Moreover, the discontinuity is infinite because $\lim\limits_{x \to 2}\frac{1-2z}{z^2-z-6}=-\infty $ .
