[amp_mcq option1=”$$\left| z \right| < {5 \over 6}$$" option2="$$\left| z \right| > {5 \over 6}$$” option3=”$${5 \over 6} < \left| z \right| < {6 \over 5}$$" option4="$${6 \over 5} < \left| z \right| < \infty $$" correct="option3"]
The correct answer is $\boxed{{5 \over 6} < \left| z \right| < {6 \over 5}}$.
The region of convergence of the z-transform of a sequence is the set of all values of $z$ for which the z-transform converges. The z-transform of the sequence ${\left( {{5 \over 6}} \right)^n}u\left( n \right) – {\left( {{6 \over 5}} \right)^n}u\left( { – n – 1} \right)$ is given by
$$Z\left{ {\left( {{5 \over 6}} \right)^n}u\left( n \right) – {\left( {{6 \over 5}} \right)^n}u\left( { – n – 1} \right) \right} = {1 \over 1 – {5 \over 6}z^{-1}} – {1 \over 1 – {6 \over 5}z}$$
The z-transform converges if and only if the denominator does not have any poles in the region of interest. The denominator has poles at $z = {6 \over 5}$ and $z = {5 \over 6}$. Therefore, the region of convergence is the set of all values of $z$ such that $z \neq {6 \over 5}$ and $z \neq {5 \over 6}$. This can be written as $${5 \over 6} < \left| z \right| < {6 \over 5}$$