The locus of the point z satisfying argleft( | vedclass

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(A) Given $arg\left( \frac{z - 1}{z + 1} \right) = k$. Let $z = x + iy$. Then $arg\left( \frac{(x - 1) + iy}{(x + 1) + iy} \right) = k$. Using the pro

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Circle with centre on $y$-axis

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(A) Given $arg\left( \frac{z - 1}{z + 1} ight) = k$. Let $z = x + iy$. Then $arg\left( \frac{(x - 1) + iy}{(x + 1) + iy} ight) = k$. Using the property $arg(z_1/z_2) = arg(z_1) - arg(z_2)$,we get $arg((x - 1) + iy) - arg((x + 1) + iy) = k$. This implies $	an^{-1}\left( \frac{y}{x - 1} ight) - 	an^{-1}\left( \frac{y}{x + 1} ight) = k$. Using $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left( \frac{A - B}{1 + AB} ight)$,we have $	an^{-1}\left( \frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + \frac{y^2}{x^2 - 1}} ight) = k$. Simplifying the expression inside: $\frac{y(x + 1) - y(x - 1)}{x^2 - 1 + y^2} = 	an k$. This simplifies to $\frac{2y}{x^2 + y^2 - 1} = 	an k$,or $x^2 + y^2 - 1 = 2y \cot k$. Rearranging gives $x^2 + y^2 - 2y \cot k - 1 = 0$. This is the equation of a circle with centre $(0, \cot k)$,which lies on the $y$-axis.

The locus of the point $z$ satisfying $arg\left( \frac{z - 1}{z + 1} \right) = k$ (where $k$ is non-zero) is

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