The value of - - arg z- arg -z-2- arg-z-argz- - is equal to - x-iy-i-1-x-y-0

The correct option is B π Since x,y gt;0 therefore arg(z) gt;0 Simplifying we get √((arg(z)+π (z) 2π)( π)) =√(( π)( π)) #160; ...

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Standard XII

Mathematics

Properties of Modulus

The value of ...

Question

The value of $∣ ∣ \sqrt{{a r g z + a r g (- ¯ z) - 2 π} {a r g (- z) + a r g (¯ z)}} ∣ ∣$ is equal to $(\forall x + i y, i = \sqrt{-} 1, x, y > 0)$

- π

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π

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Solution

The correct option is B $π$
Since $x, y > 0$ therefore $a r g (z) > 0$
Simplifying we get
$\sqrt{(a r g (z) + π - arg (z) - 2 π) (- π)}$
$= \sqrt{(- π) (- π)}$
$= π$

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Similar questions

Let

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

be any complex number,

a, b, ϵ R

and

\sqrt{- 1} = i .

Let

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

where

- 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}

Let

z

w

be non-zero complex numbers such that they have equal modulus values and

a r g z - a r g ¯ w = π,

then z equals

For

z \neq 0

, define

log z = log | z | + i (a r g z)

where

- π < a r g (z) \leq π

i.e.

a r g (z)

stands for the principal argument of

z

Then

z log (e^{x + i y})

equals :

For

z \neq 0

, define

log z = log | z | + i (a r g z)

where

- π < a r g (z) \leq π

i.e.

a r g (z)

stands for the principal argument of

z

log (- i)

equals :

Q. If

z = x + i y

such that

| z + 1 | = | z - 1 |

and

a r g (\frac{z - 1}{z + 1}) = \frac{π}{4}

, then

Q. If

| z - i | = 1

and

a r g (z) = θ, θ \in (0, \frac{π}{2}),

the the value of

cot θ - \frac{2}{z}

is equal to

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The value of ∣∣√{argz+arg(−¯z)−2π}{arg(−z)+arg(¯z)}∣∣ is equal to (∀x+iy,i=√−1,x,y>0)

The correct option is B πSince x,y>0 therefore arg(z)>0Simplifying we get√(arg(z)+π−arg(z)−2π)(−π)=√(−π)(−π) =π

The value of $∣ ∣ \sqrt{{a r g z + a r g (- ¯ z) - 2 π} {a r g (- z) + a r g (¯ z)}} ∣ ∣$ is equal to $(\forall x + i y, i = \sqrt{-} 1, x, y > 0)$

The correct option is B $π$
Since $x, y > 0$ therefore $a r g (z) > 0$
Simplifying we get
$\sqrt{(a r g (z) + π - arg (z) - 2 π) (- π)}$
$= \sqrt{(- π) (- π)}$
$= π$