Q. 164.5( 6 Votes )

# z_{1} and z_{2} are two complex numbers such that |z_{1}| = |z_{2}| and arg (z_{1}) + arg (z_{2}) = π, then show that

Answer :

Let z_{1} = |z_{1}| (cos θ_{1} + I sin θ_{1}) and z_{2} = |z_{2}| (cos θ_{2} + I sin θ_{2})

Given that |z_{1}| = |z_{2}|

And arg (z_{1}) + arg (z_{2}) = π

⇒ θ_{1} + θ_{2} = π

⇒ θ_{1} = π – θ_{2}

Now, z_{1} = |z_{2}| (cos (π - θ_{2}) + I sin (π - θ_{2}))

⇒ z_{1} = |z_{2}| (-cos θ_{2} + I sin θ_{2})

⇒ z_{1} = -|z_{2}| (cos θ_{2} – I sin θ_{2})

⇒ z_{1} = - [|z_{2}| (cos θ_{2} – I sin θ_{2})]

∴ z_{1} = -z̅ _{2}

Hence proved.

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