If |z|=1 and w=frac{z-1}{z+1} (where z neq -1 | vedclass

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(A) Given $w = \frac{z-1}{z+1}$.Since $|z|=1$,we can write $z = x+iy$ where $x^2+y^2=1$.Then $w = \frac{(x-1)+iy}{(x+1)+iy}$.To find the real part,mul

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(A) Given $w = \frac{z-1}{z+1}$.Since $|z|=1$,we can write $z = x+iy$ where $x^2+y^2=1$.Then $w = \frac{(x-1)+iy}{(x+1)+iy}$.To find the real part,multiply the numerator and denominator by the conjugate of the denominator: $(x+1)-iy$.$w = \frac{((x-1)+iy)((x+1)-iy)}{(x+1)^2+y^2} = \frac{(x^2-1) + y^2 + i(y(x+1) - y(x-1))}{(x+1)^2+y^2}$.$w = \frac{(x^2+y^2-1) + 2iy}{(x+1)^2+y^2}$.Since $|z|=1$,we have $x^2+y^2=1$,so $x^2+y^2-1=0$.Therefore,$\operatorname{Re}(w) = \frac{0}{(x+1)^2+y^2} = 0$.

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

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