If |z| = 1 and omega = frac{z - 1}{z + 1} (wh | vedclass

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(A) Given $|z| = 1$,let $z = x + iy$,so $x^2 + y^2 = 1$.$\omega = \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.To find the real part,multi

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(A) Given $|z| = 1$,let $z = x + iy$,so $x^2 + y^2 = 1$.$\omega = \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.To find the real part,multiply the numerator and denominator by the conjugate of the denominator $(x + 1) - iy$:$\omega = \frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 - 1) + i(y(x + 1) - y(x - 1)) + y^2}{(x + 1)^2 + y^2}$.$\omega = \frac{(x^2 + y^2 - 1) + i(xy + y - xy + y)}{(x + 1)^2 + y^2} = \frac{(1 - 1) + 2iy}{(x + 1)^2 + y^2} = \frac{2iy}{(x + 1)^2 + y^2}$.Since the real part is $0$,$	ext{Re}(\omega) = 0$.

If $|z| = 1$ and $\omega = \frac{z - 1}{z + 1}$ (where $z \neq -1$),then $\text{Re}(\omega)$ is

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