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

We have, |z|=1.Let z=eiθ, where θ∈R and i=−1.Then, ω=z−1z+1=eiθ−1eiθ+1=itan⁡θ2∴ Re⁡(ω)=0

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

We have, |z|=1.Let z=eiθ, where θ∈R and i=−1.Then, ω=z−1z+1=eiθ−1eiθ+1=itan⁡θ2∴ Re⁡(ω)=0

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

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

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

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

answer is A.

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Detailed Solution

We have, $| z | = 1$ .

Let $z = e^{i θ}$ , where $θ \in R$ and $i = \sqrt{- 1}$ .

Then, $ω = \frac{z - 1}{z + 1} = \frac{e^{i θ} - 1}{e^{i θ} + 1} = i \tan (\frac{θ}{2})$

$∴ Re (ω) = 0$

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

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