Let O be the origin,the point A be z_1 = sqrt | vedclass

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

Question

(D) Given $z_1 = \sqrt{3} + 2\sqrt{2}i$. The modulus is $|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$.Given $\sqrt{3}|z_2|

Vedclass · Accepted Answer

$ABO$ is an obtuse angled isosceles triangle

Vedclass · Answer

(D) Given $z_1 = \sqrt{3} + 2\sqrt{2}i$. The modulus is $|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$.Given $\sqrt{3}|z_2| = |z_1|$,so $|z_2| = \frac{|z_1|}{\sqrt{3}} = \sqrt{\frac{11}{3}}$.Given $\arg(z_2) - \arg(z_1) = \frac{\pi}{6}$,the angle $\angle AOB = \frac{\pi}{6}$.The area of $	riangle ABO = \frac{1}{2} |z_1| |z_2| \sin(\angle AOB) = \frac{1}{2} \cdot \sqrt{11} \cdot \sqrt{\frac{11}{3}} \cdot \sin(\frac{\pi}{6}) = \frac{1}{2} \cdot \frac{11}{\sqrt{3}} \cdot \frac{1}{2} = \frac{11}{4\sqrt{3}}$.Using the Law of Cosines in $	riangle ABO$,$AB^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(\frac{\pi}{6}) = 11 + \frac{11}{3} - 2 \cdot \sqrt{11} \cdot \sqrt{\frac{11}{3}} \cdot \frac{\sqrt{3}}{2} = \frac{44}{3} - 11 = \frac{11}{3}$.Since $|z_2|^2 = \frac{11}{3}$ and $AB^2 = \frac{11}{3}$,we have $|z_2| = AB$. Thus,$	riangle ABO$ is an isosceles triangle with $\angle OAB = \angle AOB = \frac{\pi}{6}$.The third angle $\angle ABO = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \frac{2\pi}{3}$.Since $\angle ABO = \frac{2\pi}{3} > \frac{\pi}{2}$,the triangle is obtuse angled isosceles.

Let $O$ be the origin,the point $A$ be $z_1 = \sqrt{3} + 2\sqrt{2}i$,and the point $B(z_2)$ be such that $\sqrt{3}|z_2| = |z_1|$ and $\arg(z_2) = \arg(z_1) + \frac{\pi}{6}$. Then:

Similar Questions

If $|z_1| = 2$,$|z_2| = 3$,$|z_3| = 4$ and $|2z_1 + 3z_2 + 4z_3| = 9$,then the value of $|8z_2z_3 + 27z_3z_1 + 64z_1z_2|$ is equal to:

If $z=x+iy$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,then the locus of $z$ is

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then

The locus of a point on the Argand plane represented by the complex number $z$, when $z$ satisfies the condition $\left|\frac{z-1+i}{z+1-i}\right|=\left|\operatorname{Re}\left(\frac{z-1+i}{z+1-i}\right)\right|$ is

Let $x_1, x_2, x_3, x_4$ be the roots of the equation $4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0$. If $(4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^2) = \frac{125}{16}m$,then the value of $m$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module