If
$$
z+\frac{1}{z}=1 \Rightarrow \frac{z^2+1}{z}=1 \Rightarrow z^2-z+1=0
$$

So, $\quad z=\frac{+1 \pm \sqrt{1-4}}{2} \quad z=\frac{1 \pm \sqrt{3} i}{2}$
$\Rightarrow \quad z=-\omega$ and $-\omega^2$
Now, for $z=-\omega$
$$
\begin{aligned}
& =\frac{\left(z^{20}+1\right)\left(z^{40}+1\right)\left(z^{60}+1\right)}{z^{60}} \\
& =\frac{\left.\left[(-\omega)^{20}+1\right]\left[(-\omega)^{40}+1\right)\right]\left[(-\omega)^{60}+1\right]}{(-\omega)^{60}} \\
& =\frac{\left(\omega^{20}+1\right)\left(\omega^{40}+1\right)\left(\omega^{60}+1\right)}{\omega^{60}}
\end{aligned}
$$
$$
=\frac{\left(\omega^2+1\right)(\omega+1)(1+1)}{1}
$$
$$
\left[\begin{array}{c}
\because \omega^{3 k}=1 \\
\omega^{3 k+1}=\omega \\
\text { and } \omega^{3 k+2}=\omega^2
\end{array}\right]
$$


$$
\begin{aligned}
& =2\left(\omega^3+\omega^2+\omega+1\right) \\
& =2(1)=2 \quad\left[\because \omega^2+\omega+1=0\right] \\
& \text { for } z=-\omega^2 \\
& \frac{\left(z^{20}+1\right)\left(z^{40}+1\right)\left(z^{60}+1\right)}{z^{60}} \\
& =\frac{\left(\left(-\omega^2\right)^{20}+1\right)\left[\left(-\omega^2\right)^{40}+1\right]\left[\left(-\omega^2\right)^{60}+1\right]}{\left(-\omega^2\right)^{60}} \\
& =\frac{\left(\omega^{40}+1\right)\left(\omega^{80}+1\right)\left(\omega^{120}+1\right)}{\omega^{120}} \\
& =\frac{(\omega+1)\left(\omega^2+1\right)(1+1)}{1} \quad\left[\begin{array}{c}
\because \omega^{3 k}=1 \\
\omega^{3 k+1}=\omega \\
\text { and } \omega^{3 k+2}=\omega^2
\end{array}\right] \\
& =2\left(\omega^3+\omega+\omega^2+1\right) \\
& =2(1)=2 \quad\left[\because \omega^2+\omega+1=0\right] \\
&
\end{aligned}
$$