First slide

Algebra of complex numbers

Question

Suppose $z$ is a complex number such that $z \neq - 1$ , $| z | = 1,$ and $\arg (z) =$ $θ$ Let $ω$ $= \frac{z (1 - \bar{z})}{\bar{z} (1 + z)}$ ,than $Re (ω)$ ) is equal to

Moderate

A

$1 + \cos (θ / 2)$
B

$1 - \sin (θ / 2)$
C

$- 2 \sin^{2} (θ / 2)$
D

$2 \cos^{2} (θ / 2)$

Solution

$\begin{array}{l} ω = \frac{z - z \bar{z}}{\bar{z} + \bar{z} z} = \frac{z - 1}{1 / z + 1} = \frac{(z - 1) z}{z + 1} [∵ | z | = 1] \\ = \frac{z^{2} - 1}{z + 1} - \frac{z - 1}{z + 1} = z - 1 + \frac{1 - z}{1 + z} \\ Re (ω) = Re (z) - 1 + Re (\frac{1 - z}{1 + z}) \end{array}$
But $Re (\frac{1 - z}{1 + z}) = \frac{1}{2} (\frac{1 - z}{1 + z} + \frac{1 - z}{1 + z})$
$= \frac{1}{2} (\frac{1 - z}{1 + z} + \frac{z - 1}{z + 1}) = 0$
$∴ Re (ω) = \cos θ - 1 = - 2 \sin^{2} (θ / 2)$

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