The number of complex numbers $z$, satisfying $|z|=1$ and $\left|\dfrac{z}{\bar z}+\dfrac{\bar z}{z}\right|=1$, is:
Step-by-Step Solution
Key Concept: With $|z|=1$ we have $\bar z = 1/z$, so $\dfrac{z}{\bar z}+\dfrac{\bar z}{z}=z^{2}+\bar z^{2}=2(x^{2}-y^{2})$ where $z=x+iy$. Combine with $x^{2}+y^{2}=1$.
Let $z=x+iy$ with $x^{2}+y^{2}=1$. Then
$$\frac{z}{\bar z}+\frac{\bar z}{z} = z^{2}+\bar z^{2} = (x+iy)^{2}+(x-iy)^{2} = 2(x^{2}-y^{2}).$$
So $|2(x^{2}-y^{2})|=1 \Longrightarrow |x^{2}-y^{2}|=\tfrac{1}{2}.$
\textbf{Case A: } $x^{2}-y^{2}=\tfrac{1}{2}$ together with $x^{2}+y^{2}=1$ gives $x^{2}=\tfrac{3}{4},\ y^{2}=\tfrac{1}{4}$, i.e. $4$ solutions.
\textbf{Case B: } $x^{2}-y^{2}=-\tfrac{1}{2}$ gives $x^{2}=\tfrac{1}{4},\ y^{2}=\tfrac{3}{4}$, another $4$ solutions.
Total: $\boxed{8}$.
Correct Answer: 2