If - z -1 and w -z-1-z-1 where -z-1- then Re w isA- -1-z-1-2B- -z-z-1- 1-z-1-2-2-z-1-2D-

| z|=1⇒| x+iy|=1⇒ x^2+y^2=1 w=z 1/z+1=(x 1)+iy/(x+1)+iy×(x+1) iy/(x+1) iy =x^2+y^2 1/(x+1)^2=2iy/(x+1)^2+y^2=2iy/(x+1)^2+y^2 (∵ x^2+y^2=1) ∴ Re(w)=0 ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

If | z |=1 an...

Question

If $∣ z ∣= 1$ $a n d^{w = \frac{z - 1}{z + 1} (w h e r e z \neq - 1), t h e n R e (w)}$ is

0

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Solution

The correct option is A $0$
$∣ z ∣= 1 \Rightarrow∣ x + i y ∣= 1 \Rightarrow x^{2} + y^{2} = 1$
$w = \frac{z - 1}{z + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y}$
$= \frac{x^{2} + y^{2} - 1}{(x + 1)^{2}} = \frac{2 i y}{(x + 1)^{2} + y^{2}} = \frac{2 i y}{(x + 1)^{2} + y^{2}}$
$(∵ x^{2} + y^{2} = 1)$
$∴ R e (w) = 0$

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