First slide

Algebra of complex numbers

Question

If $|z| = 1, z \neq 1,$ then value of $a r g (\frac{1}{1 - z})$ cannot exceed

Moderate

A

$π / 2$
B

$π$
C

$3 π / 2$
D

$2 π$

Solution

As $| z | = 1, z \neq 1, z = \cos θ + i \sin θ_{,} - π < θ \leq π, θ \neq 0 .$ Now
$ω = \frac{1}{1 - z} = \frac{1}{1 - \cos θ - i \sin θ}$
$\begin{array}{l} = \frac{(1 - \cos θ) + i \sin θ}{(1 - \cos θ)^{2} + \sin^{2} θ} = \frac{(1 - \cos θ) + i \sin θ}{2 (1 - \cos θ)} \\ = \frac{1}{2} + \frac{i}{2} \cot (\frac{θ}{2}) = \frac{1}{2} + \frac{i}{2} \tan (\frac{π}{2} - \frac{θ}{2}) \end{array}$
This shows that $w$ lies on the line $x = 1 / 2$ and $- π / 2 < \arg (ω) < π / 2, Arg (ω) \neq 0,$ The maximum value of $A r g (w)$ is never attained.

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