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Geometrical Representation of Argument and Modulus

If arg z > 0,...

Question

If arg (z) > 0, then arg (-z) -arg(z) is equal to

π

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- π

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\frac{π}{2}

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\frac{- π}{2}

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Solution

The correct option is B $- π$
$a r g (- z) - a r g (z)$
$= a r g (- z) + a r g (¯ ¯ ¯ z)$
Now $a r g (z) > 0$
$∴ a r g (- z) + a r g (¯ ¯ ¯ z) = - π$

Suggest Corrections

Similar questions

a r g (z) < 0

a r g (- z) - a r g (z)

| 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,

z = \frac{cos θ + i sin θ}{cos θ - i sin θ}

\frac{π}{4} < 0 < \frac{π}{2}

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