least value of floor of $z_{1}+z_{2}$

Question

If $z_{1},z_{2},z_{3},z_{4}$ are the roots of the equation

$z^4+z^3+z^2+z+1=0.$ Then the least value of

$\lfloor |z_{1}+z_{2}|\rfloor +1$ is

Try: From given equation $z^4+z^3+z^2+z+1=0$

$$\frac{z^5-1}{z-1}=0(z-1\neq 0\Rightarrow z\neq 1.)$$

So $$z^5=1\Rightarrow z = (1)^{\frac{1}{5}}=\bigg(\cos 2k\pi+i\sin 2k\pi\bigg)^{\frac{1}{5}}$$

So $$z=e^{i\frac{2k\pi}{5}} = \cos \frac{2k\pi}{5}+i\sin \frac{2k\pi}{5}.$$

For $k=0,1,2,3,4$

So $$z_{i}\in \cos \frac{2k\pi}{5}+i\sin \frac{2k\pi}{5}$$ for $i=1,2,3,4,5.$

Now i did not understand how i find least of $z_{1}+z_{2}.$

could some help me to solve it, thanks

Answer 1

Note that minimum value of $|z_1+z_2|$ is the minimum magnitude of the vector that can be obtained by the sum of vectors with angle $2\pi /5$ and $4\pi/5$ between them, all having magnitude of $1$ (by geometrical interpretation of $n^{th}$ roots of $1$).

$$|z_1+z_2|=\sqrt{1+1+2\cos\theta}$$ $\cos\theta=0.309$ OR $-0.809$ (corresponding to $72$ and $144$ degrees respectively). So, $$\text{min}|z_1+z_2|=\sqrt{0.381}\implies \text{ans}=0+1=\boxed{1}.$$

Answer 2

Let $\alpha=e^{i\frac{2\pi}{5}}$, then each root of the given equation $z^4+\dotsb+1=0$ is a power of $\alpha$. Let $z_1=\alpha^k$ and $z_2=\alpha^j$, where $k \leq j$ and $k,j \in \{1,2,3,4\}$.

Let us first consider $$|z_1+z_2|=|\alpha^k||1+\alpha^{j-k}|=|1+\alpha^{j-k}|=\sqrt{2}\sqrt{1+\cos\left(\frac{2\pi(j-k)}{5}\right)}=2\left|\cos\left(\frac{2\pi(j-k)}{10}\right)\right|$$

Thus we want to minimize $\color{blue}{\left\lfloor 2\left|\cos\left(\frac{2\pi(j-k)}{10}\right)\right|\right\rfloor +1}$.

Observe that the only possible values of $\left\lfloor 2\left|\cos\left(\frac{2\pi(j-k)}{10}\right)\right|\right\rfloor$ are $0,1,2$ ($\because 0 \leq |\cos \theta| \leq 1$).

We can show that this could be zero. Because $j-k \in \{0,1,2,3,4\}$, therefore $j-k$ can take the value $2$. In which case, (using the decreasing nature of the cosine function) we have $$\frac{4\pi}{10} > \frac{\pi}{3} \implies \cos\left(\frac{4\pi}{10}\right) < \cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$

Thus $\left\lfloor 2\left|\cos\left(\frac{2\pi(j-k)}{10}\right)\right|\right\rfloor$ can take the value $0$. Thus the minimum value of the given expression is $\color{red}{1}$.

Answer 3

$$\frac{z^5-1}{z-1}=0(z-1\neq 0\Rightarrow z\neq 1.)$$

So $$z^5=1\Rightarrow z = (1)^{\frac{1}{5}}=\bigg(\cos 2k\pi+i\sin 2k\pi\bigg)^{\frac{1}{5}}$$

So $$z=e^{i\frac{2k\pi}{5}} = \cos \frac{2k\pi}{5}+i\sin \frac{2k\pi}{5}.$$

For $k=0,1,2,3,4$

Not quite. Don't forget that $z \neq 1$, so $k \neq 0$.

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$|z_1| = |z_2| = 1$ because they're roots of unity. Therefore $|z_1 + z_2| \le 2$ with equality only when $z_1 = z_2$. Also $0 \le |z_1 + z_2|$ with equality only when $z_1 = -z_2$.

Thus immediately we see that $\lfloor |z_{1}+z_{2}|\rfloor$ is $0$ or $1$.

To see geometrically whether it must be $1$, consider the Argand diagram. If you draw a circle of unit radius centred on the origin, and then a second circle of unit radius centred on $z_1$ (on the circumference of the first circle), by the standard compass-and-ruler construction of a hexagon a third of the circumference of the second circle lies inside the first circle. Therefore the question is whether any of the other roots lie in that third when we translate the second circle back to the first circle, and the answer is obviously yes, since they're spaced every fifth of the circumference.

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Alternatively, taking a more algebraic approach: since the coefficients are real, roots occur in conjugate pairs. It suffices to find a root whose real part is in the range $(-\tfrac12, \tfrac12)$.