If arg (z)

Question

If arg (z) < 0 , then arg (-z) - arg (z) equals (A) π (B) -π (C) -π/2 (D) π/2

Tardigrade Admin · Accepted Answer

Since, arg (z) < 0 ⇒ arg (z)=-θ <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/0aa5d869bfc9728387d6ecd76fd35c1f-.png" /> ⇒ z=r cos (-θ)+i sin (-θ) =r( cos θ-i sin θ) and -z=-r[ cos θ-i sin θ] =r[ cos (π-θ)+i sin (π-θ)] ∴ arg (-z)=π-θ Thus, arg (-z)- arg (z) =π-θ-(-θ)=π

Q. If $a r g (z) < 0$ , then $a r g (- z) - a r g (z)$ equals

$π$

$- π$

$- π /2$

$π /2$

Since, $ar g (z) < 0$
$\Rightarrow ar g (z) = - θ$

$\Rightarrow z = r cos (- θ) + i sin (- θ)$
$= r (cos θ - i sin θ)$
and $- z = - r [cos θ - i sin θ]$
$= r [cos (π - θ) + i sin (π - θ)]$
$∴ ar g (- z) = π - θ$
Thus, $ar g (- z) - ar g (z)$
$= π - θ - (- θ) = π$

Q. If arg(z)<0 , then arg(−z)−arg(z) equals

π

−π

−π/2

π/2

Since, arg(z)<0 ⇒arg(z)=−θ ⇒z=rcos(−θ)+isin(−θ) =r(cosθ−isinθ) and −z=−r[cosθ−isinθ] =r[cos(π−θ)+isin(π−θ)] ∴arg(−z)=π−θ Thus, arg(−z)−arg(z) =π−θ−(−θ)=π

Q. If $a r g (z) < 0$ , then $a r g (- z) - a r g (z)$ equals

$π$

$- π$

$- π /2$

$π /2$

Since, $ar g (z) < 0$
$\Rightarrow ar g (z) = - θ$

$\Rightarrow z = r cos (- θ) + i sin (- θ)$
$= r (cos θ - i sin θ)$
and $- z = - r [cos θ - i sin θ]$
$= r [cos (π - θ) + i sin (π - θ)]$
$∴ ar g (- z) = π - θ$
Thus, $ar g (- z) - ar g (z)$
$= π - θ - (- θ) = π$