Statement -I : If z1 and z2 are two nonzero complex numbers such that z1+z2=z1+z2 then ag z1- ag z2 is π2.Statement-II : z1 and z2 are two complex numbers such that z1z2=1 and arg z1-arg z2 is π2then z1 z2=-i

Let arg⁡z1=θ and arg⁡z2=ϕ given thatz1+z2=z1+z2 On squaring both sides, we get z1+z22=z12+z22+2z1z2⇒z12+z22+2z1z2cos⁡(θ−ϕ)=z12+z22+2z1z2⇒ cos⁡(θ−ϕ)=1∴ θ−ϕ=0 or θ=ϕ∴ arg⁡z1=arg⁡z2 z1z2=1 and arg z1-arg z2=π2 ⇒-arg z1+arg z2=-π2 ⇒arg z1+arg z2=-π2 ⇒argz1 z2=-π2 ⇒z1 z2=z1 z2e-iπ2=-i

Statement -I : If z1 and z2 are two nonzero complex numbers such that z1+z2=z1+z2 then ag z1- ag z2 is π2.Statement-II : z1 and z2 are two complex numbers such that z1z2=1 and arg z1-arg z2 is π2then z1 z2=-i

Let arg⁡z1=θ and arg⁡z2=ϕ given thatz1+z2=z1+z2 On squaring both sides, we get z1+z22=z12+z22+2z1z2⇒z12+z22+2z1z2cos⁡(θ−ϕ)=z12+z22+2z1z2⇒ cos⁡(θ−ϕ)=1∴ θ−ϕ=0 or θ=ϕ∴ arg⁡z1=arg⁡z2 z1z2=1 and arg z1-arg z2=π2 ⇒-arg z1+arg z2=-π2 ⇒arg z1+arg z2=-π2 ⇒argz1 z2=-π2 ⇒z1 z2=z1 z2e-iπ2=-i

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Statement -I : If z₁and z₂are two nonzero complex numbers such that $|z_{1} + z_{2}| = |z_{1}| + |z_{2}|$ then ag z₁- ag z₂ is $\frac{π}{2}$ .
Statement-II : z₁and z₂are two complex numbers such that $|z_{1} z_{2}| = 1 and \arg z_{1} - \arg z_{2}$ is $\frac{π}{2}$ then $z_{1} z_{2} = - i$

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Both I and II are true

Neither I nor II are true

Only I is true

Only II is true

answer is B.

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Detailed Solution

$Let \arg (z_{1}) = θ and \arg (z_{2}) = ϕ g i v e n t h a t |z_{1} + z_{2}| = |z_{1}| + |z_{2}| On squaring both sides, we get \begin{array}{l} {|z_{1} + z_{2}|}^{2} = {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \\ \Rightarrow {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \cos (θ - ϕ) \\ = {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \\ \Rightarrow \cos (θ - ϕ) = 1 \\ ∴ θ - ϕ = 0 or θ = ϕ \\ ∴ \arg (z_{1}) = \arg (z_{2}) \end{array}$

$|z_{1} z_{2}| = 1 and \arg z_{1} - \arg z_{2} = \frac{π}{2} \Rightarrow - \arg z_{1} + \arg z_{2} = - \frac{π}{2} \Rightarrow \arg z_{1} + \arg z_{2} = - \frac{π}{2} \Rightarrow \arg z_{1} z_{2} = - \frac{π}{2} \Rightarrow z_{1} z_{2} = |z_{1} z_{2}| e^{- i \frac{π}{2}} = - i$