How complexity does a number achieve? #2

Write $z = r e^{i \theta}$ . Since $|z|=1$ , then $r=1$ , hence $z = e^{i \theta}$ . Now, we have, $\frac{z}{\bar{z}} + \frac{\bar{z}}{z} = 2 \left(\frac{e^{i 2\theta} + e^{-i 2\theta}}{2} \right) = 2 \cos(2\theta)$ then, $\left| \frac{z}{\bar{z}} + \frac{\bar{z}}{z}\right | = 1$ imply $\left| \cos(2\theta) \right| =\frac{1}{2}$

The equation have 2 solutions for each quadrant, so in total we have 8 solutions.

Let $z = a + bi$ , then $|z| = |a+bi| = 1 \quad \Rightarrow \sqrt{a^2+b^2} = 1\quad \Rightarrow a^2+b^2 = 1$

Now, we have:

$\left| \dfrac {z}{\bar{z}} + \dfrac{\bar{z}} {z} \right| = \left| \dfrac {a+bi}{a-bi} + \dfrac {a-bi}{a+bi} \right| = \left| \dfrac {(a+bi)^2+(a-bi)^2}{(a-bi)(a+bi)} \right| \\ = \left| \dfrac {a^2+2abi-b^2+a^2-2abi-b^2}{a^2+b^2} \right| =\left| 2(a^2-b^2) \right| = 1 \\ \Rightarrow \left| a^2-b^2 \right| = \frac {1}{2} \quad \Rightarrow a^2-b^2 = \pm \frac {1}{2}$

$\Rightarrow \begin{cases} a^2+b^2 = 1 \\ a^2-b^2 = \pm \frac {1}{2} \end{cases} \quad \Rightarrow 2a^2 = \begin{cases} \frac {3}{2} \\ \frac {1}{2} \end{cases} \quad \Rightarrow a = \begin{cases} \pm \frac {\sqrt{3}}{2} \\ \pm \frac {1}{2} \end{cases}$

$\Rightarrow \begin{array} {llll} z = -\frac{\sqrt{3}}{2}-\frac{1}{2}i & z = -\frac{\sqrt{3}}{2}+\frac{1}{2}i & z = -\frac{1}{2} -\frac{\sqrt{3}}{2}i & z = -\frac{1}{2} +\frac{\sqrt{3}}{2}i \\ z = \frac{1}{2} -\frac{\sqrt{3}}{2}i & z = \frac{1}{2} +\frac{\sqrt{3}}{2}i & z = \frac{\sqrt{3}}{2}-\frac{1}{2}i & z = \frac{\sqrt{3}}{2}+\frac{1}{2}i \end{array}$

Therefore, there are $\boxed{8}$ such complex number $z$ .

Replace conjugate of z by 1/z and replace mod Square by x*(conjugate x). You get an eight degree polynomial.

Let $z = a + bi$ ; from $|z| = 1$ , we get that $a^{2} + b^{2} = 1$

Now, $\frac{z}{\overline{z}} + \frac{\overline{z}}{z} = \frac{2a^{2} - 2b^{2}}{a^{2} + b^{2}} = 2(a^{2} - b^{2})$ ; thus, we can write:

$|\frac{z}{\overline{z}} + \frac{\overline{z}}{z}| = |2(a^{2} - b^{2})|$

Thus, we have two cases:

Case 1:

$\begin{cases} a^{2} + b^{2} = 1 \\ a^{2} - b^{2} = \frac{1}{2} \end{cases}$

This yields four solutions:

$\frac{1 + \sqrt{3}i}{2}, \frac{1 - \sqrt{3}i}{2}, \frac{-1 + \sqrt{3}i}{2}, \frac{-1 - \sqrt{3}i}{2}$

Case 2:

$\begin{cases} a^{2} + b^{2} = 1 \\ - a^{2} + b^{2} = \frac{1}{2} \end{cases}$

This yields four solutions:

$\frac{\sqrt{3} + i}{2}, \frac{\sqrt{3} - i}{2}, \frac{-\sqrt{3} + i}{2}, \frac{-\sqrt{3} - i}{2}$

Thus, we have a total of eight solutions.