Z-Transform

System is all-pass filter.
For digital all-pass filter, condition is
\text{zero}=\frac{1}{\text{pole}^{*}}\; \; ...(i)
By given transfer function,
\text{zero}=\alpha
\text{Pole}=-0.5
using condition(i),
\alpha =\frac{1}{-0.5}=-2

P(Z)=H(Z)H\left ( \frac{1}{Z} \right )
(i) h(n) is real. Som p(n) will be also real
(ii) P(z)=P(z^{-1})
From (i) : if z_1 is a zero of P(z), then z_1^* will be also a zero of P(z).
From (ii): If z_1 is a zero of P(z), then \frac{1}{z_1} will be also a zero of P(z).
So, the 4 zeros are,
\begin{aligned} z_1&= \frac{1}{2}+\frac{1}{2}j\\ z_2&= z_1^*=\frac{1}{2}-\frac{1}{2}j\\ z_3&=\frac{1}{z_1}=\frac{1}{\frac{1}{2}-\frac{1}{2}j}=1-j \\ z_4&=\left ( \frac{1}{z_1} \right )^*=z_3^*=1+j \end{aligned}

The AOC should include unit circle to make the system stable. From the given pole pattern it is clear that, to make the system stable, the ROC should be finite circular strip or finite annular strip and hence the impulse response of the system should be also two-sided.

\begin{array}{l} \begin{aligned} \text { Given, } & x[n]=(2.0)^{\ln 1},-\infty \lt n \lt +\infty \\ &=2^{n}(n]+\left(\frac{1}{2}\right)^{n} u(-n-1) \end{aligned} \\ \text{ROC}:|z| \gt 2 \text { and }|z| \lt \frac{1}{2}\\ \therefore \text{No Roc} \end{array}

Given,
\begin{aligned} x[n] &=\delta[n-3]+2 \delta[n-5] \\ X(z) &=z^{3}+2 z^{5} \\ X(-z) &=(-z)^{-3}+2(-z)^{-5} \\ Y(z) &=X(-z)=-z^{-3}-2 z^{-5} \\ &=-\left[z^{-3}+2 z^{-5}\right] \\ y[n] &=-x[n] \end{aligned}

\begin{array}{l} \text { Given, } x[n]=a^{n} u[n]+b^{n} u[n] \\ \text { Also given, } \quad \begin{aligned} 0 &<|a|<|b|<1 \\ & A O C=(|z|>|a|) \text { and }(|z|>|b|) \\ & R O C=|z|>|b| \end{aligned} \end{array}

\begin{aligned} \frac{Y(z)}{X(z)}&=\frac{\frac{5}{3}[1-z]}{z^{2}-z+\frac{2}{9}} \\ \Rightarrow \quad \eta (z)&=\frac{5}{3} \frac{(1-z)}{\left(z-\frac{1}{3}\right)\left(z-\frac{2}{3}\right)}\\ \eta(z) &=\frac{5 z}{z-\frac{1}{3}}-\frac{5 z}{z-\frac{2}{3}} \\ y(n) &=5\left(\frac{1}{3}\right)^{n} u(n)-5\left(\frac{2}{3}\right)^{n} u(n) \end{aligned}

Poles are at z=\pm 2
Since x[n] is absolutely summable hence, DTFT exists which implies ROC must include unit circle.
\Rightarrow x(t) is a non-causal signal.

\begin{aligned} y(n)&=\sum_{i=0}^{3} \alpha_{i} x(n-i) \\ &=\alpha_{0}x[n]+\alpha_{1} x[n-1]+\alpha_{2} x[n-2]+\alpha_{3} x[n-3] \\ \Rightarrow Y(z)&=\alpha_{0} X(z)+\alpha_{1} X(z) z^{-1}+\alpha_{2} X(z) z^{-2} \\ & +\alpha_{3} X(z) z^{-3} \\ H(z)&=\alpha_{0}+\alpha_{1} z^{-1}+\alpha_{2} z^{-2}+\alpha_{3} z^{-3} \\ \text {Null at }&\text{zero frequency } \\ \Rightarrow \quad z&=\left.e^{j \omega}\right|_{\omega=0}\\ 0 &= \alpha_{0}+\alpha_{1}+\alpha_{2}+\alpha_{3} \end{aligned}
Option (A) satisfies the condition

\begin{aligned} f(n)&=\sum_{m=0}^{n} x[m]=x[n] * u[n] \\ y(z)&=X(z) \cdot\left(\frac{z}{z-1}\right)\\ \text{Given} \quad Y(z)&=\frac{2}{z(z-1)^{2}}=X(z) \cdot \frac{z}{z-1} \\ \Rightarrow \quad A(z)&=\frac{2 \cdot z^{-2}}{z-1}=2 z^{-3}\left(\frac{z}{z-1}\right) \\ \Rightarrow \quad x[n]&=2 u[n-3] \\ \text{Therefore, } x[2]&=0 \end{aligned}