The correct option is A z_2=kz_1^ 1(k gt;0)arg(z̅_̅1̅)=arg(z̅_̅2̅) ⇒ arg(z_1)+arg(z_2)=0 ⇒ z_1 z_2 =k ...

If arg z1=z2,...

Question

If arg $(_{1}) = a r g (_{2})$ , then

z_{2} = k z_{1}^{- 1} (k > 0)

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z_{2} = k z_{1} (k > 0)

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| z_{2} | = |_{1} |

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Solution

The correct option is A $z_{2} = k z_{1}^{- 1} (k > 0)$
$a r g (_{1}) = a r g (_{2}) \Rightarrow a r g (z_{1}) + a r g (z_{2}) = 0 \Rightarrow z_{1} z_{2} = k$

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Similar questions

(Z_{1} + Z_{2}) = 0

| z_{1} | = | z_{2} | = 1.

If | $z_{1}$ + $z_{2}$ | =| $z_{1}$ - $z_{2}$ | , then arg $z_{1}$ - arg $z_{2}$ =

z_{1}

z_{2}

| z_{1} + z_{2} | = | z_{1} | + | z_{2} |

z_{1} -

z_{2}

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a r g z_{1} & a r g z_{2} & a r g z_{3} a r g z_{2} & a r g z_{3} & a r g z_{1} a r g z_{3} & a r g z_{1} & a r g z_{2} \end{matrix} ∣ ∣ ∣ ∣

Δ

z_{1}

z_{2}

| z_{1} | = | z_{2} |

z_{1} +

z_{2} = 0

z_{1}

z_{2}