If Z neq pm 1 is a complex number and operato | vedclass

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(A) 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 denomina

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$x^2+y^2-2y-1=0$

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(A) 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 $(x+1) - iy$,we get:$\frac{Z-1}{Z+1} = \frac{((x-1) + iy)((x+1) - iy)}{(x+1)^2 + y^2} = \frac{(x^2 - 1 + y^2) + i(y(x+1) - y(x-1))}{(x+1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + 2iy}{(x+1)^2 + y^2}$.Given $\operatorname{Arg}\left(\frac{Z-1}{Z+1}ight) = \frac{\pi}{4}$,we have $	an\left(\frac{\pi}{4}ight) = \frac{	ext{Imaginary part}}{	ext{Real part}} = \frac{2y}{x^2 + y^2 - 1}$.Since $	an\left(\frac{\pi}{4}ight) = 1$,we have $1 = \frac{2y}{x^2 + y^2 - 1}$.This simplifies to $x^2 + y^2 - 1 = 2y$,or $x^2 + y^2 - 2y - 1 = 0$.

If $Z \neq \pm 1$ is a complex number and $\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right)=\frac{\pi}{4}$,then the locus of $Z$ in the Argand plane is

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