Roots of a Complex Number expression If $z_1$, $z_2$, $z_3$, $z_4$ are roots of the equation $z^4+z^3+z^2+z+1=0$,
then what is the least value of $\lfloor mod(z_1 + z_2)\rfloor + 1$? 
($\lfloor.\rfloor$ denotes Greatest Integer Function)
 A: The roots are
$$w_k = e^{\frac{2\pi k}{5}i} \quad(k = 1,2,3,4)$$
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&= \min\left\lfloor\left|e^{\frac{2\pi m}{5}i}+e^{\frac{2\pi n}{5}i}\right|\right\rfloor+1\\
&= \min \left\lfloor\left|e^{\frac{2\pi m}{5}i}\right|\left|1+e^{\frac{2\pi (n-m)}{5}i}\right|\right\rfloor+1\\
&=\min\left\lfloor\left|1+e^{\frac{2\pi (n-m)}{5}i}\right|\right\rfloor+1\\
&=\min\left\lfloor\sqrt{\left[1+\cos{\frac{2\pi(n-m)}{5}}\right]^2 + \left[\sin{\frac{2\pi(n-m)}{5}}\right]^2}\right\rfloor+1\\
&=\min\left\lfloor\sqrt{2+2\cos{\frac{2\pi(n-m)}{5}}}\right\rfloor+1\\
&=\min\left\lfloor\sqrt{4\cos^2{\frac{2\pi(n-m)}{10}}}\right\rfloor+1\\
&=\min\left\lfloor2\left|\cos{\frac{2\pi(n-m)}{10}}\right|\right\rfloor+1\\
\end{align*}$$
(The identity $\left|1+e^{i\theta}\right| \equiv 2\left|\cos\frac \theta2\right|$ also has a geometric interpretation)
The least value is when the $\cos$ is closest to zero, when $(n-m) = 2$ (or 3 or -2 or -3):
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&=\left\lfloor2\left|\cos{\frac{2\pi\cdot2}{10}}\right|\right\rfloor+1\\
&=\left\lfloor2\left|\cos{\frac{2\pi}{5}}\right|\right\rfloor+1\\
\end{align*}$$
Without knowing the actual value of $\cos \frac{2\pi}{5}$, note that
$$\frac{2\pi}{6}<\frac{2\pi}{5}<\frac{2\pi}{4}\\
\frac12 > \cos \frac{2\pi}{5} > 0\\
1 > 2 \left|\cos \frac{2\pi}{5}\right| > 0\\
\left\lfloor2 \left|\cos \frac{2\pi}{5}\right|\right\rfloor = 0$$
So the least value is
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&= 0+1\\
&= 1
\end{align*}$$
A: $$z^4+z^3+z^2+z+1=\frac{z^5-1}{z-1}=0$$
$$\therefore z^5=1\text{ and } z\ne1$$
These solutions can be plotted on an Argand diagram which more easily shows that the solution is given by the two complex conjugates with the smallest magnitude real parts - $e^{\frac{2i\pi}5}$ and $e^{\frac{-2i\pi}5}$. Adding these values gives
$$e^{\frac{2i\pi}5}+e^{-\frac{2i\pi}5}=\cos{\left({\frac{2\pi}5}\right)}+i\sin{\left({\frac{2\pi}5}\right)}+\cos{\left({\frac{2\pi}5}\right)}-i\sin{\left({\frac{2\pi}5}\right)}=2\cos{\left({\frac{2\pi}5}\right)}$$
So the smallest value is given by
$$\lfloor z_1+z_2\rfloor+1=\left\lfloor 2\cos{\left({\frac{2\pi}5}\right)}\right\rfloor+1=\lfloor 0.618...\rfloor+1=1$$
