The locus of the points z which satisfy the c | vedclass

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(B) Let $z = x + iy$. Then $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.Multiplying the numerator and denominator by the conjugate of the

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$A$ circle

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(B) Let $z = x + iy$. Then $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.Multiplying the numerator and denominator by the conjugate of the denominator,we get $\frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2}$.Given $	ext{arg} \left( \frac{z - 1}{z + 1} ight) = \frac{\pi}{3}$,we have $	an^{-1} \left( \frac{2y}{x^2 + y^2 - 1} ight) = \frac{\pi}{3}$.This implies $\frac{2y}{x^2 + y^2 - 1} = 	an \frac{\pi}{3} = \sqrt{3}$.Rearranging the terms,we get $x^2 + y^2 - 1 = \frac{2}{\sqrt{3}}y$,or $x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0$.This is the equation of a circle.

The locus of the points $z$ which satisfy the condition $\text{arg} \left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3}$ is

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