The locus of the complex number z such that a | vedclass

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(A) Let $z = x + iy$.Given $\arg \left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}$.Using the property $\arg \left(\frac{z_1}{z_2}\right) = \arg(z_1) - \ar

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a circle

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(A) Let $z = x + iy$.Given $\arg \left(\frac{z-2}{z+2}ight) = \frac{\pi}{3}$.Using the property $\arg \left(\frac{z_1}{z_2}ight) = \arg(z_1) - \arg(z_2)$,we have:$\arg(z-2) - \arg(z+2) = \frac{\pi}{3}$$\arg((x-2) + iy) - \arg((x+2) + iy) = \frac{\pi}{3}$$	an^{-1}\left(\frac{y}{x-2}ight) - 	an^{-1}\left(\frac{y}{x+2}ight) = \frac{\pi}{3}$Applying the formula $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$:$	an^{-1}\left[\frac{\frac{y}{x-2} - \frac{y}{x+2}}{1 + \left(\frac{y}{x-2}ight)\left(\frac{y}{x+2}ight)}ight] = \frac{\pi}{3}$$\frac{\frac{y(x+2) - y(x-2)}{x^2-4}}{\frac{x^2-4+y^2}{x^2-4}} = 	an\left(\frac{\pi}{3}ight)$$\frac{4y}{x^2+y^2-4} = \sqrt{3}$$x^2 + y^2 - \frac{4}{\sqrt{3}}y - 4 = 0$This is the equation of a circle.

The locus of the complex number $z$ such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$ is:

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