If -z-1 and - -z-1-z-1 where- z - -1- then Re - is

Since,|z|=1 and omega=z 1/z+1 ⇒ z 1=ω z+ω⇒ z=1+ω/1 ω⇒ |z|=|1+ω|/|1 ω| ⇒ |1 ω|=|1+ω| [∴ |z|=1] On squaring both sides,we get 1+|ω|^2 2Re(ω)=1+|ω|^2+2Re (ω) ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

If |z|=1 and ...

Question

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

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

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$∣ ∣ \begin{matrix} \frac{1}{z + 1} \end{matrix} ∣ ∣ . \frac{1}{| z + 1 |^{2}}$

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

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Solution

The correct option is A
0

Since, $| z | = 1 a n d o m e g a = \frac{z - 1}{z + 1} \Rightarrow z - 1 = ω z + ω \Rightarrow z = \frac{1 + ω}{1 - ω} \Rightarrow | z | = \frac{| 1 + ω |}{| 1 - ω |} \Rightarrow | 1 - ω | = | 1 + ω | [∴ | z | = 1]$
On squaring both sides,we get
$1 + | ω |^{2} - 2 R e (ω) = 1 + | ω |^{2} + 2 R e (ω)$
using $| z_{1} \pm z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} \pm 2 R e ({¯ z}_{1} z_{2})] \Rightarrow 4 R e (ω) = 0 \Rightarrow R e (ω) = 0$

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

The correct option is A 0 Since,|z|=1 and omega=z−1z+1⇒z−1=ωz+ω⇒z=1+ω1−ω⇒|z|=|1+ω||1−ω|⇒|1−ω|=|1+ω| [∴|z|=1] On squaring both sides,we get 1+|ω|2−2Re(ω)=1+|ω|2+2Re(ω) using |z1±z2|2=|z1|2+|z2|2±2Re(¯z1z2)]⇒4Re(ω)=0⇒Re(ω)=0

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