The area of the region enclosed by the locus | vedclass

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(B) The locus of $z$ satisfying $	ext{Arg}\left(\frac{z+i}{z-i}ight) = \frac{2\pi}{3}$ is an arc of a circle.Let $z = x + iy$. The points $A(0, -1)

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$\frac{4\pi}{9} - \frac{1}{\sqrt{3}}$

Vedclass · Answer

(B) The locus of $z$ satisfying $	ext{Arg}\left(\frac{z+i}{z-i}ight) = \frac{2\pi}{3}$ is an arc of a circle.Let $z = x + iy$. The points $A(0, -1)$ and $B(0, 1)$ are fixed.The angle subtended by the chord $AB$ at any point $z$ on the arc is $\alpha = \frac{2\pi}{3}$.The center of the circle lies on the real axis at $C\left(-\cot\left(\frac{2\pi}{3}ight), 0ight) = C\left(-\frac{1}{\sqrt{3}}, 0ight)$.The radius $R$ is given by $R = \frac{AB}{2\sin(\alpha)} = \frac{2}{2\sin(2\pi/3)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$.The area of the circular segment is given by $\frac{1}{2}R^2(	heta - \sin 	heta)$,where $	heta$ is the central angle.The central angle $	heta = 2(\pi - \alpha) = 2(\pi - 2\pi/3) = 2\pi/3$.Area $= \frac{1}{2} \left(\frac{2}{\sqrt{3}}ight)^2 \left(\frac{2\pi}{3} - \sin\left(\frac{2\pi}{3}ight)ight) = \frac{1}{2} \cdot \frac{4}{3} \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}ight) = \frac{2}{3} \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}ight) = \frac{4\pi}{9} - \frac{1}{\sqrt{3}}$.

The area of the region enclosed by the locus of $z$ given by $\text{Arg}(z + i) - \text{Arg}(z - i) = \frac{2\pi}{3}$ and the imaginary axis is:

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