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If z_{1} and z_{2} are two complex numbers such that | z_{1}

Question

If $z_{1}$ and $z_{2}$ are two complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣$ and $a r g (z_{1}) + r a g (z_{2}) = π$ , then show that $z_{1} = - \overset{z}{ˉ}_{2}$ .

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Camila Discussed

z_{1}

and

z_{2}

are two complex numbers such that

∣ z_{1} ∣ = ∣ z_{2} ∣

and

a r g (z_{1}) + r a g (z_{2}) = π

, then show that

z_{1} = - \overset{z}{ˉ}_{2}

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Text solutionVerified

Let

z_{1} = r_{1} (cos θ_{1} + i s in θ_{1})

and

z_{2} = r_{2} (cos θ_{2} + i s in θ_{2})

Then,

∣ z_{1} ∣ = r_{1}, a r g (z_{1}) = θ_{1}, ∣ z_{2} ∣ = r_{2}

and

a r g (z_{2}) = θ_{2}

It is given that

∣ z_{2} ∣ = ∣ z_{1} ∣

and

a r g (z_{1}) + a r g (z_{2}) = π

\Rightarrow r_{1} = r_{2}

and

θ_{1} + θ_{2} = π

\Rightarrow r_{1} = r_{2}

and

θ_{1} = π - θ_{2}

∴ z_{1} = r_{1} (cos θ_{1} + i s in θ_{1})

\Rightarrow z_{1} = r_{2} {cos (π - θ_{2}) + i s in (π - θ_{2})} = r_{2} (- cos θ_{2} + i s in θ_{2})

\Rightarrow z_{1} = - r_{2} (cos θ_{2} - i s in θ_{2}) = - \overset{z}{ˉ}_{2}

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Question Text	If $z_{1}$ and $z_{2}$ are two complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣$ and $a r g (z_{1}) + r a g (z_{2}) = π$ , then show that $z_{1} = - \overset{z}{ˉ}_{2}$ .
Answer Type	Text solution:1
Upvotes	150