If | z | = 1 and = z - 1 z + 1 ( where z - 1 ) , then Re ( ) is [2003]

Question

If $| z | = 1$ and $ω = \frac{z - 1}{z + 1} (where z \neq - 1)$ , then $Re (ω)$ is [2003]

Smart Achievers · Accepted Answer

(1)

$Given that | z | = 1 and ω = \frac{z - 1}{z + 1} (z \neq - 1)$

$Now we know that z \bar{z} = | z |^{2}$

$\Rightarrow z \bar{z} = 1 (for | z | = 1)$

$∴$ $ω = (\frac{z - 1}{z + 1}) \times (\frac{\bar{z} + 1}{\bar{z} + 1})$

$= \frac{z \bar{z} + z - \bar{z} - 1}{z \bar{z} + z + \bar{z} + 1}$ $= \frac{2 i y}{2 + 2 x}$

$[∵ z \bar{z} = 1 and taking z = x + i y so that z + \bar{z} = 2 x and z - \bar{z} = 2 i y]$

$\Rightarrow Re (ω) = 0$

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