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Properties of Argument

If argz <0, t...

Question

If arg(z) <0, then arg (-z)-arg(z)=

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-π

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-π/2

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π/2

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Solution

The correct option is A
π

Let arg(z) = $α$ < 0, Then z = $r e^{i α}$

Now -z = $e^{π i} . r e^{i α}$ = $r e^{i (π + α)}$

$\Rightarrow a r g (- z)$ = $π + α ∴ a r g (- z) - a r g (z)$ = $π$

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Similar questions

| π - arg z | < \frac{π}{4},

(z)

z = 1 + i b = (a, b)

a, b, ϵ R

\sqrt{- 1} = i .

z \neq 0 + 0 i, a r g z = {tan}^{- 1} (\frac{I m z}{R e z})

- π < a r g z \leq π

a r g (¯ z) + a r g (- z) = {\begin{matrix} π, i f a r g (z) < 0 - π, i f a r g (z) > 0 \end{matrix}

z

w

a r g z - a r g ¯ w = π,

a r g z = \frac{π}{6}

z =

| Z | = 2

a r g (Z) = \frac{π}{4}

Z

z = \frac{π}{4} (1 + i)^{4} (\frac{1 - π i}{π + i} + \frac{π - i}{1 + π i})

(\frac{| z |}{a r g (z)})

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