Assertion :Statement -1: If z1 and z2 are two complex numbers such that ∣z1∣=∣z2∣+∣z1−z2∣, then Im(z2z1)=0 Reason :Statement -2: arg(z)=0⇒z is purely real.

Given, ∣z1 ∣=∣z2 ∣+∣z1−z2 ∣⇒∣z1−z2 ∣=∣z1 ∣−∣z2 ∣⇒∣z1−z2 ∣2=(∣z1 ∣−∣z2 ∣)2(if∣z1 ∣≥∣z2 ∣)⇒∣z1 ∣2−2∣z1 ∣∣z2 ∣cos(θ1−θ2)+∣z2∣2=∣z1 ∣2+∣z2 ∣2−2∣z1 ∣∣z2 ∣⇒cos(θ1−θ2)=1⇒θ1−θ2=2nπ,n∈INow z2z1=∣∣z2z1 ∣∣[cos(θ1−θ2)+isin(θ1−θ2) ]=∣∣z2z1 ∣∣⇒Im(z2z1 )=0Hence, Statement-1 is True. Hence, option A is correct.

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Assertion :Statement -1: If z_1 and z_2 are two complex number

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Assertion :Statement -1: If $z_{1}$ and $z_{2}$ are two complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ + ∣ z_{1} - z_{2} ∣$ , then $I m (\frac{z _{1}}{z _{2}}) = 0$
Reason :Statement -2: $a r g (z) = 0 \Rightarrow z$ is purely real.

Statement -1 is true, Statement -2 is true ; Statement -2 is a correct explanation for Statement -1

Statement-1 is true, Statement-2 is true ; Statement-2 is not a correct explanation for Statement-1

Statement -1 is true, Statement -2 is false

Statement -1 is false, Statement -2 is true

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Henry Discussed

Assertion :Statement -1: If

z_{1}

and

z_{2}

are two complex numbers such that

∣ z_{1} ∣ = ∣ z_{2} ∣ + ∣ z_{1} - z_{2} ∣

, then

I m (\frac{z _{1}}{z _{2}}) = 0

Reason :Statement -2:

a r g (z) = 0 \Rightarrow z

is purely real.

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Text solutionVerified

Given,

∣ z_{1} ∣ = ∣ z_{2} ∣ + ∣ z_{1} - z_{2} ∣

\Rightarrow ∣ z_{1} - z_{2} ∣ = ∣ z_{1} ∣ - ∣ z_{2} ∣

\Rightarrow ∣ z_{1} - z_{2} ∣^{2} = (∣ z_{1} ∣ - ∣ z_{2} ∣)^{2} (i f ∣ z_{1} ∣ \geq ∣ z_{2} ∣)

\Rightarrow ∣ z_{1} ∣^{2} - 2 ∣ z_{1} ∣ ∣ z_{2} ∣ cos (θ_{1} - θ_{2}) + ∣ z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} - 2 ∣ z_{1} ∣ ∣ z_{2} ∣

\Rightarrow cos (θ_{1} - θ_{2}) = 1 \Rightarrow θ_{1} - θ_{2} = 2 nπ, n \in I

Now

\frac{z _{1}}{z _{2}} = ∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣ [cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2})] = ∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣

\Rightarrow I m (\frac{z _{1}}{z _{2}}) = 0

Hence, Statement-1 is True.
Hence, option A is correct.

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Question Text	Assertion :Statement -1: If $z_{1}$ and $z_{2}$ are two complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ + ∣ z_{1} - z_{2} ∣$ , then $I m (\frac{z _{1}}{z _{2}}) = 0$ Reason :Statement -2: $a r g (z) = 0 \Rightarrow z$ is purely real.
Answer Type	Text solution:1
Upvotes	150

Assertion :Statement -1: If z1​ and z2​ are two complex numbers such that ∣z1​∣=∣z2​∣+∣z1​−z2​∣, then Im(z2​z1​​)=0 Reason :Statement -2: arg(z)=0⇒z is purely real.

Text solutionVerified

Practice questions from similar books

Assertion :Statement -1: If $z_{1}$ and $z_{2}$ are two complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ + ∣ z_{1} - z_{2} ∣$ , then $I m (\frac{z _{1}}{z _{2}}) = 0$
Reason :Statement -2: $a r g (z) = 0 \Rightarrow z$ is purely real.