= Solution
For a pure <qubit> $a|0\rangle+b|1\rangle$, the <Bloch vector> is $(2\operatorname{Re}(a^*b),2\operatorname{Im}(a^*b),|a|^2-|b|^2)$. Substitution gives
$$
\boxed{z=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta),\qquad
\overline z=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,-\cos\theta).}
$$
Both lie on the unit <Bloch sphere>, at the same azimuth and opposite heights. They are reflections across its equatorial plane, not antipodal vectors. Choose $\theta=\pi/3$ and $\varphi=\pi/4$, so their coordinates are $(\sqrt6/4,\sqrt6/4,\pm1/2)$. The following original sketch shows those vectors and their common meridian:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-58-bloch-vectors.png]
{title=Bloch vectors at theta pi/3 and azimuth pi/4, reflected across the equatorial plane}
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