JEE Advanced 2023 Paper 2 · Q11 · Complex Numbers

Question

Let $A_1, A_2, A_3, \ldots, A_8$ be the vertices of a regular octagon that lie on a circle of radius $2$. Let $P$ be a point on the circle and let $PA_i$ denote the distance between the points $P$ and $A_i$ for $i = 1, 2, \ldots, 8$. If $P$ varies over the circle, then the maximum value of the product $PA_1 \cdot PA_2 \cdots PA_8$ is ___.

SolveKar AI · Accepted Answer

Place the octagon with vertices at the $8$th roots of $2^8 = 256$, i.e. $A_k = 2\omega^k$ where $\omega = e^{2\pi i/8}$. For a point $P = z$ on the circle of radius $2$, the product $\prod_{k=1}^{8}|z - 2\omega^k| = |z^8 - 2^8|$. The maximum of $|z^8 - 256|$ for $|z| = 2$ is achieved when $z^8 = -256$, giving $|{-256} - 256| = 512$. So $\max\prod PA_i = 512$.

JEE Advanced 2023 Paper 2 · Q11 · Complex Numbers

Solution