Let $P(z)$, $R(z^4)$ and $Q(z^2)$ be three points in the Argand plane such that $PR + RQ = PQ$; ($z\neq 0,1$). Then $z$ lies on
Step-by-Step Solution
Key Concept: $PR+RQ=PQ$ means $R$ lies on segment $PQ$ (collinearity with $R$ between $P$ and $Q$). So $z,z^2,z^4$ are collinear with $z^4$ between $z$ and $z^2$ (or another ordering).
$R$ between $P$ and $Q$: $\text{Im}\left(\frac{z^4-z}{z^2-z}\right)=0$. $\frac{z^4-z}{z^2-z}=\frac{z(z^3-1)}{z(z-1)}=\frac{z^3-1}{z-1}=z^2+z+1$. For this to be real, $\text{Im}(z^2+z+1)=0$. Let $z=x+iy$: $\text{Im}(z^2+z)=2xy+y=y(2x+1)=0$. Either $y=0$ (real axis) or $x=-1/2$ (vertical line). Since $z\neq 0,1$ and $z$ on $x=-1/2$: a line parallel to $y$-axis.
Correct Answer: 4