If $\omega$ is an imaginary fifth root of unity, then $\log_2 \begin{vmatrix} 1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\ \end{vmatrix}$ = 
If $\omega$ is an imaginary fifth root of unity, then $$\log_2 \begin{vmatrix}
1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\
\end{vmatrix} =$$

My approach : 
$$\omega^5 = 1 \\ \implies 1+\omega +\omega^2 +\omega^3 + \omega^4 =0$$
Therefore, \begin{align}\log_2 |1+\omega +\omega^2+ \omega^3 -\frac{1}{\omega}| &=\log_2 |1+\omega +\omega^2+ \omega^3 -\omega^4|\\& =\log_2|-2\omega^4|\\ &=\log_2 2 +\log_2 \omega^4 \end{align}
Now how to solve further; please suggest. 

Since
$$
\begin{align}
1+\omega+\omega^2+\omega^3+\omega^4
&=\frac{1-\omega^5}{1-\omega}\\[3pt]
&=0
\end{align}
$$
we have
$$
\begin{align}
1+\omega+\omega^2+\omega^3-\frac1\omega
&=-\omega^4-\frac1\omega\\
&=-\frac{\omega^5+1}\omega\\
&=-\frac2\omega
\end{align}
$$
Therefore
$$
\begin{align}
\log_2\left|1+\omega+\omega^2+\omega^3-\frac1\omega\right|
&=\log_2\left|-\frac2\omega\right|\\
&=\log_2(2)\\[6pt]
&=1
\end{align}
$$
