If $\left| z \right| = 1$ and $w = \frac{{z - 1}}{{z + 1}}$ (where $z \ne - 1$), then ${\rm{Re}}\left( w \right)$ is

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

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

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

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Medium

1 of 1

Application +4 | -1

\left| z \right| = 1

and

w = \frac{{z - 1}}{{z + 1}}

(where

z \ne - 1

), then

\mathrm{Re}\left( w \right)

View answer and explanation

Correct option: A

Option A (0) is correct because the real part of

w

is 0, which is derived using the property that

|z| = 1

and

w = \frac{{z - 1}}{{z + 1}}

Practice All Chapter MCQs →

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\left|\sum\limits_{r=1}^n z_r\right|

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\alpha, \beta

\gamma

|\alpha-\beta|

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