GATE Electronics and Communications (EC) 2019 Solved Paper

let $\mathit{H}\left(\mathit{z}\right)=\frac{\left(\mathit{z}-\mathit{a}\right)\left(\mathit{z}-\mathit{b}\right)}{\left(\mathit{z}-\mathit{c}\right)\left(\mathit{z}-\mathit{d}\right)}$
$\therefore \mathit{H}\left(\frac{1}{\mathit{z}}\right)=\frac{\left(\frac{1}{\mathit{z}}-\mathit{a}\right)\left(\frac{1}{\mathit{z}}-\mathit{b}\right)}{\left(\frac{1}{\mathit{z}}-\mathit{c}\right)\left(\frac{1}{\mathit{z}}-\mathit{d}\right)}$
$\therefore \mathit{H}\left(\frac{1}{\mathit{z}}\right)=\frac{\left(\mathit{z}-\frac{1}{\mathit{a}}\right)\left(\mathit{z}-\frac{1}{\mathit{b}}\right)}{\left(\mathit{z}-\frac{1}{\mathit{c}}\right)\left(\mathit{z}-\frac{1}{\mathit{d}}\right)}$
$\therefore$ zeros are $\mathit{a},\mathit{b},\frac{1}{\mathit{a}},\frac{1}{\mathit{b}}$
given zero is $\mathit{a}=\frac{1}{2}+\frac{1}{2}\mathit{j}$
as $\mathit{h}\left(\mathit{n}\right)$ is real valued signal another zero must be complex conjugate of this
$\therefore \mathit{b}=\frac{1}{2}-\frac{1}{2}\mathit{j}$
Now ${\mathit{z}}_3=\frac{1}{\mathit{a}}=\frac{1}{\frac{1}{2}+\frac{1}{2}\mathit{j}}$
$=\frac{2}{1+\mathit{j}}$
$=\frac{2\left(1-\mathit{j}\right)}{2}$
${\mathit{z}}_3=1-\mathit{j}$
as $\mathit{h}\left(\mathit{n}\right)$ is real valued signal another zero must be complex conjugate of this ${\mathit{z}}_4=1+\mathit{j}$