Let A_1, A_2, A_3, ldots, A_8 be the vertices | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the vertices of the regular octagon be represented by complex numbers $z_k = 2e^{i(	heta_0 + \frac{2\pi(k-1)}{8})}$ for $k=1, 2, \ldots, 8$.

Vedclass · Accepted Answer

$512$

Vedclass · Answer

(C) Let the vertices of the regular octagon be represented by complex numbers $z_k = 2e^{i(	heta_0 + \frac{2\pi(k-1)}{8})}$ for $k=1, 2, \ldots, 8$. Without loss of generality,let $	heta_0 = 0$. The vertices are the roots of the equation $z^8 - 2^8 = 0$.Thus,$z^8 - 2^8 = \prod_{k=1}^8 (z - A_k)$.Let $P$ be represented by the complex number $z = 2e^{i	heta}$. The distance $PA_k = |z - A_k|$.The product is $\prod_{k=1}^8 PA_k = |\prod_{k=1}^8 (z - A_k)| = |z^8 - 2^8|$.Substituting $z = 2e^{i	heta}$,we get $|(2e^{i	heta})^8 - 2^8| = |2^8 e^{i8	heta} - 2^8| = 2^8 |e^{i8	heta} - 1|$.Using the identity $|e^{i\phi} - 1| = |\cos \phi + i \sin \phi - 1| = \sqrt{(\cos \phi - 1)^2 + \sin^2 \phi} = \sqrt{2 - 2\cos \phi} = 2|\sin(\frac{\phi}{2})|$.Here,$\phi = 8	heta$,so the product is $2^8 \cdot 2|\sin(4	heta)| = 2^9 |\sin(4	heta)| = 512 |\sin(4	heta)|$.The maximum value of $|\sin(4	heta)|$ is $1$.Therefore,the maximum value of the product is $512 	imes 1 = 512$.

Similar Questions

If the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ lies on the locus of $z$ satisfying the inequality $\left|\frac{z+3i}{3z+i}\right| < 1$,then the interval in which $k$ lies is

If the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,then the locus of $z=x+iy$ is:

If $A, B, C$ are represented by $3 + 4i, 5 - 2i, -1 + 16i$,then $A, B, C$ are

Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then,the maximum value of $|z-(6+8i)|$ is equal to:

If $z$ is a complex number,then the minimum value of $|z| + |z - 1|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module