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(D) 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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$x^2 + y^2 - 2y = 1$

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(D) 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 $(x + 1) - iy$:$\frac{z - 1}{z + 1} = \frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 - 1 + y^2) + i(y(x + 1) - y(x - 1))}{(x + 1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2}$.Given that the amplitude is $\frac{\pi}{4}$,we have $	an(\frac{\pi}{4}) = \frac{	ext{Imaginary part}}{	ext{Real part}} = \frac{2y}{x^2 + y^2 - 1}$.Since $	an(\frac{\pi}{4}) = 1$,we get $1 = \frac{2y}{x^2 + y^2 - 1}$.Thus,$x^2 + y^2 - 1 = 2y$,which simplifies to $x^2 + y^2 - 2y = 1$.

If a complex number $z = x + iy$ is taken such that the amplitude of the fraction $\frac{z - 1}{z + 1}$ is always $\frac{\pi}{4}$,then:

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