Let z 1 and z 2 be two complex numbers such that z ¯ 1 = i z ¯ 2 and arg z 1 z ¯ 2 = π , then the argument of z 1 is

z ¯ 1 = i z ¯ 2 ⇒ z 1 = - i z 2 ⇒ i z 1 = z 2 ⇒ e i π 2 z 1 = z 2 Given, arg z 1 z ¯ 2 = π Let z 1 = r e i θ ⇒ z 2 ¯ = r e - i π 2 + θ So arg e i 2 θ + π 2 = π ⇒ 2 θ + π 2 = π ⇒ θ = π 4

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Mathematics - Complex Number Question with Solution | TestHub

MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Easy2 minPYQ_2022

MathematicsEasysingle choice

Let $z_{1}$ and $z_{2}$ be two complex numbers such that ${\bar{z}}_{1} = i {\bar{z}}_{2}$ and $\arg (\frac{z_{1}}{{\bar{z}}_{2}}) = π$ , then the argument of $z_{1}$ is

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Solution:

${\bar{z}}_{1} = i {\bar{z}}_{2} \Rightarrow z_{1} = - i z_{2} \Rightarrow i z_{1} = z_{2} \Rightarrow e^{\frac{i π}{2}} z_{1} = z_{2}$

Given, $\arg (\frac{z_{1}}{{\bar{z}}_{2}}) = π$

Let $z_{1} = r e^{i θ} \Rightarrow \bar{z_{2}} = r e^{- i (\frac{π}{2} + θ)}$

So $\arg (e^{i (2 θ + \frac{π}{2})}) = π \Rightarrow 2 θ + \frac{π}{2} = π$
$\Rightarrow θ = \frac{π}{4}$

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2022

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