If z = x + iy and argleft( frac{z - 2}{z + 2} | vedclass

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(B) Given $\arg\left( \frac{z - 2}{z + 2} \right) = \frac{\pi}{6}$.Substitute $z = x + iy$:$\arg\left( \frac{(x - 2) + iy}{(x + 2) + iy} \right) = \fr

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$A$ circle

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(B) Given $\arg\left( \frac{z - 2}{z + 2} ight) = \frac{\pi}{6}$.Substitute $z = x + iy$:$\arg\left( \frac{(x - 2) + iy}{(x + 2) + iy} ight) = \frac{\pi}{6}$.Using the property $\arg(z_1/z_2) = \arg(z_1) - \arg(z_2)$:$	an^{-1}\left( \frac{y}{x - 2} ight) - 	an^{-1}\left( \frac{y}{x + 2} ight) = \frac{\pi}{6}$.Using 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 + \frac{y^2}{x^2 - 4}} ight) = \frac{\pi}{6}$.$\frac{y(x + 2) - y(x - 2)}{x^2 - 4 + y^2} = 	an\left( \frac{\pi}{6} ight) = \frac{1}{\sqrt{3}}$.$\frac{4y}{x^2 + y^2 - 4} = \frac{1}{\sqrt{3}}$.$x^2 + y^2 - 4\sqrt{3}y - 4 = 0$.This is the equation of a circle.

If $z = x + iy$ and $\arg\left( \frac{z - 2}{z + 2} \right) = \frac{\pi}{6}$,then the locus of $z$ is

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