If |z|=1 and ω=z−1z+1 (where z≠−1), then Re(ω) is

|z|=1⇒|x+iy|=1⇒x2+y2=1 ω=z−1z+1=(x−1)+iy(x+1)+iy×(x+1)−iy(x+1)−iy=(x2+y2−1)(x+1)2+y2+2iy(x+1)2+y2=2iy(x+1)2+y2

If |z|=1 and ω=z−1z+1 (where z≠−1), then Re(ω) is

|z|=1⇒|x+iy|=1⇒x2+y2=1 ω=z−1z+1=(x−1)+iy(x+1)+iy×(x+1)−iy(x+1)−iy=(x2+y2−1)(x+1)2+y2+2iy(x+1)2+y2=2iy(x+1)2+y2

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$I f | z | = 1 a n d ω = \frac{z - 1}{z + 1} (w h e r e z \neq - 1), t h e n Re (ω) i s$

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$| \frac{z}{z + 1} | . \frac{1}{{| z + 1 |}^{2}}$

$- \frac{1}{{| z + 1 |}^{2}}$

$\frac{\sqrt{2}}{{| z + 1 |}^{2}}$

answer is A.

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Detailed Solution

$| z | = 1 \Rightarrow | x + i y | = 1 \Rightarrow x^{2} + y^{2} = 1 ω = \frac{z - 1}{z + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} = \frac{(x^{2} + y^{2} - 1)}{{(x + 1)}^{2} + y^{2}} + \frac{2 i y}{{(x + 1)}^{2} + y^{2}} = \frac{2 i y}{{(x + 1)}^{2} + y^{2}}$