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. 
 A: 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
$$
\begin{align}
\log_2\left|1+\omega+\omega^2+\omega^3-\frac1\omega\right|
&=\log_2\left|-\frac2\omega\right|\\
&=\log_2(2)\\[6pt]
&=1
\end{align}
$$
A: Notice:
$$\omega^5=1\Longleftrightarrow$$
$$\omega^5=e^{0i}\Longleftrightarrow$$
$$\omega=\left(e^{2\pi ki}\right)^{\frac{1}{5}}\Longleftrightarrow$$
$$\omega=e^{\frac{2\pi ki}{5}}$$
With $k\in\mathbb{Z}$ and $k:0-4$

So, the solutions are:
$$\omega_0=e^{\frac{2\pi\cdot0i}{5}}=e^{\frac{0}{5}}=e^0=1$$
$$\omega_1=e^{\frac{2\pi\cdot1i}{5}}=e^{\frac{2\pi i}{5}}=e^{\frac{2\pi i}{5}}$$
$$\omega_2=e^{\frac{2\pi\cdot2i}{5}}=e^{\frac{4\pi i}{5}}=e^{\frac{4\pi i}{5}}$$
$$\omega_3=e^{\frac{2\pi\cdot3i}{5}}=e^{\frac{6\pi i}{5}}=e^{-\frac{4\pi i}{5}}$$
$$\omega_4=e^{\frac{2\pi\cdot4i}{5}}=e^{\frac{8\pi i}{5}}=e^{-\frac{2\pi i}{5}}$$
So:
$$\omega=
\begin{cases}
1\\
e^{\pm\frac{2\pi i}{5}}\\
e^{\pm\frac{4\pi i}{5}}
\end{cases}$$




*

*When $\omega=1$:
$$\log_2\left|1+1+1^2+1^3-\frac{1}{1}\right|=\log_2\left|1+1+1+1-1\right|=\log_2\left|3\right|=\log_2(3)$$

*When $\omega=e^{\pm\frac{2\pi i}{5}}$:
$$\log_2\left|1+e^{\pm\frac{2\pi i}{5}}+\left(e^{\pm\frac{2\pi i}{5}}\right)^2+\left(e^{\pm\frac{2\pi i}{5}}\right)^3-\frac{1}{e^{\pm\frac{2\pi i}{5}}}\right|=1$$

*When $\omega=e^{\pm\frac{4\pi i}{5}}$:
$$\log_2\left|1+e^{\pm\frac{4\pi i}{5}}+\left(e^{\pm\frac{4\pi i}{5}}\right)^2+\left(e^{\pm\frac{4\pi i}{5}}\right)^3-\frac{1}{e^{\pm\frac{4\pi i}{5}}}\right|=1$$
