Key Concepts and Formulas

• Complex Conjugate and Modulus: For a complex number $z = x + iy$, its conjugate is $\bar{z} = x - iy$, and its modulus is $|z| = \sqrt{x^2 + y^2}$. Also, $z\bar{z} = |z|^2$.
• Unit Circle Property: If $|z| = 1$, then $z\bar{z} = 1$, which implies $\bar{z} = \frac{1}{z}$.
• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$, and $e^{-i\theta} = \cos\theta - i\sin\theta$. Also, $e^{ix} + e^{-ix} = 2\cos x$.

Step-by-Step Solution

Step 1: Simplify the Second Condition using $|z|=1$

We are given $|z| = 1$ and $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$. We want to simplify the second condition using the fact that $|z|=1$ implies $z\bar{z}=1$, and therefore $\bar{z} = \frac{1}{z}$.

$\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$ Substitute $\bar{z} = \frac{1}{z}$ into the expression: $\left|\frac{z}{1/z}+\frac{1/z}{z}\right|=1$ $\left|z^2+\frac{1}{z^2}\right|=1$ Since $|z| = 1$, we have $z\bar{z}=1$ which implies $\bar{z} = \frac{1}{z}$. Therefore $\frac{1}{z^2} = (\frac{1}{z})^2 = (\bar{z})^2$. Then: $\left|z^2+(\bar{z})^2\right|=1$

Step 2: Express $z$ in Cartesian Form ($x+iy$) and Apply the Modulus Conditions

Let $z = x + iy$, where $x$ and $y$ are real numbers. Then $\bar{z} = x - iy$. From $|z| = 1$, we have:

$|x+iy| = 1$ $x^2 + y^2 = 1 \quad (*)$ Now, calculate $z^2$ and $(\bar{z})^2$: $z^2 = (x+iy)^2 = x^2 + 2ixy - y^2$ $(\bar{z})^2 = (x-iy)^2 = x^2 - 2ixy - y^2$ Therefore: $z^2 + (\bar{z})^2 = (x^2 + 2ixy - y^2) + (x^2 - 2ixy - y^2) = 2(x^2 - y^2)$ Substitute this into the simplified second condition: $|2(x^2 - y^2)| = 1$ $2|x^2 - y^2| = 1$ $|x^2 - y^2| = \frac{1}{2}$ This gives us two cases: Case I: $x^2 - y^2 = \frac{1}{2}$ Case II: $x^2 - y^2 = -\frac{1}{2}$

Step 3: Solve the Systems of Equations for Each Case

We solve each case in conjunction with the equation $x^2 + y^2 = 1$.

Case I: Solving $x^2+y^2=1$ and $x^2-y^2=\frac{1}{2}$

We have the system: $x^2 + y^2 = 1$ $x^2 - y^2 = \frac{1}{2}$ Adding the two equations: $2x^2 = \frac{3}{2}$ $x^2 = \frac{3}{4}$ $x = \pm \frac{\sqrt{3}}{2}$ Substituting $x^2 = \frac{3}{4}$ into $x^2 + y^2 = 1$: $\frac{3}{4} + y^2 = 1$ $y^2 = \frac{1}{4}$ $y = \pm \frac{1}{2}$ This gives us four solutions: $(\frac{\sqrt{3}}{2}, \frac{1}{2})$, $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$, $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$, $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$.

Case II: Solving $x^2+y^2=1$ and $x^2-y^2=-\frac{1}{2}$

We have the system: $x^2 + y^2 = 1$ $x^2 - y^2 = -\frac{1}{2}$ Adding the two equations: $2x^2 = \frac{1}{2}$ $x^2 = \frac{1}{4}$ $x = \pm \frac{1}{2}$ Substituting $x^2 = \frac{1}{4}$ into $x^2 + y^2 = 1$: $\frac{1}{4} + y^2 = 1$ $y^2 = \frac{3}{4}$ $y = \pm \frac{\sqrt{3}}{2}$ This gives us four solutions: $(\frac{1}{2}, \frac{\sqrt{3}}{2})$, $(\frac{1}{2}, -\frac{\sqrt{3}}{2})$, $(-\frac{1}{2}, \frac{\sqrt{3}}{2})$, $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$.

Step 4: Total Number of Solutions

In total, we have 4 solutions from Case I and 4 solutions from Case II, giving us a total of 8 solutions.

Step 5: Alternative Solution using Polar Form

Let $z = e^{i\theta}$, then $\bar{z} = e^{-i\theta}$. The condition $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$ becomes: $\left|\frac{e^{i\theta}}{e^{-i\theta}}+\frac{e^{-i\theta}}{e^{i\theta}}\right|=1$ $\left|e^{2i\theta}+e^{-2i\theta}\right|=1$ Using Euler's formula, $e^{ix} + e^{-ix} = (\cos x + i\sin x) + (\cos x - i\sin x) = 2\cos x$: $|2\cos(2\theta)|=1$ $2|\cos(2\theta)|=1$ $|\cos(2\theta)|=\frac{1}{2}$ This implies $\cos(2\theta) = \frac{1}{2}$ or $\cos(2\theta) = -\frac{1}{2}$. For $0 \le \theta < 2\pi$ (one full rotation), we need to consider $0 \le 2\theta < 4\pi$.

• If $\cos(2\theta) = \frac{1}{2}$, then $2\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$. Thus, $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$. (4 solutions)
• If $\cos(2\theta) = -\frac{1}{2}$, then $2\theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$. Thus, $\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$. (4 solutions) These 8 distinct values of $\theta$ in $[0, 2\pi)$ each correspond to a unique complex number $z$ on the unit circle.

Common Mistakes & Tips

• Remember that $|z|=1$ implies $z\bar{z}=1$, which is crucial for simplifying the given expression.
• When solving for $x$ and $y$, always consider both positive and negative roots.
• Using the polar form can sometimes lead to a more direct solution.

Summary

We used the property $|z|=1$ to simplify the given equation. Then we expressed $z$ in Cartesian form and solved the resulting system of equations. We found a total of 8 complex numbers that satisfy both conditions.

The final answer is \boxed{8}, which corresponds to option (A).