The number of complex numbers $z$ such that $|z|=1$ and $\left|\dfrac{z}{\bar{z}}+\dfrac{\bar{z}}{z}\right|=1$ is
Step-by-Step Solution
Key Concept: Simplify $z/\bar{z}+\bar{z}/z = (z^2+\bar{z}^2)/|z|^2 = 2(x^2-y^2)$ on the unit circle, then intersect $|x^2-y^2|=1/2$ with $x^2+y^2=1$.
On $|z|=1$: $\dfrac{z}{\bar{z}}+\dfrac{\bar{z}}{z}=z^2+\bar{z}^2=2(x^2-y^2)$. So the condition is $|2(x^2-y^2)|=1$, i.e. $x^2-y^2=\pm\dfrac{1}{2}$.
**With $x^2+y^2=1$:**
$x^2-y^2=\tfrac{1}{2}\Rightarrow x^2=\tfrac{3}{4},\,y^2=\tfrac{1}{4}$: 4 points.
$x^2-y^2=-\tfrac{1}{2}\Rightarrow x^2=\tfrac{1}{4},\,y^2=\tfrac{3}{4}$: 4 points.
Total: **8 complex numbers**.
Correct Answer: 2