Roots of a Complex Number expression

Question

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)

Answer 1

$$z^4+z^3+z^2+z+1=\frac{z^5-1}{z-1}=0$$ $$\therefore z^5=1\text{ and } z\ne1$$ 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 $$\lfloor z_1+z_2\rfloor+1=\left\lfloor 2\cos{\left({\frac{2\pi}5}\right)}\right\rfloor+1=\lfloor 0.618...\rfloor+1=1$$

Answer 2

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

$$\begin{align*} \min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1 &= 0+1\\ &= 1 \end{align*}$$