If z 1- z 2 are two complex numbers such that - z 1- z 2-2 and z 1- z 2-0 then z 1 z 2-

Let z_1=|z_1|(cosθ+isinθ) where θ=(z_1) and z_2=|z_2|(cosϕ+isinϕ) where ϕ=(z_2) It is given that z_2= z_1 ⇒ϕ= θ Also, nbsp;|z_1|=|z_2| z_2=|z_1|[cos( θ)+is ...

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Standard XII

Mathematics

Argument of a Complex Number

If z 1, z 2 a...

Question

If $z_{1}, z_{2}$ are two complex numbers such that $| z_{1} | = | z_{2} | = 2$ and $arg z_{1} + arg z_{2} = 0$ then $z_{1} z_{2} =$

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Solution

Let $z_{1} = | z_{1} | (cos θ + i sin θ)$ where $θ = arg (z_{1})$ and
$z_{2} = | z_{2} | (cos ϕ + i sin ϕ)$ where $ϕ = arg (z_{2})$
It is given that $arg z_{2} = - arg z_{1} \Rightarrow ϕ = - θ$ and $| z_{1} | = | z_{2} |$
So $z_{2} = | z_{1} | (cos (- θ) + i sin (- θ))$
$\Rightarrow z_{2} = | z_{1} | (cos θ - i sin θ)$
$\Rightarrow z_{2} =_{1} ∴ z_{1} z_{2} =_{2} z_{2} = | z_{2} |^{2} = 4$

$Alternate Solution$
$arg z_{1} + arg z_{2} = 0$
$\Rightarrow arg (z_{1} z_{2}) = 0$
So, $z_{1} z_{2} = k, (k > 0)$
Now, $| z_{1} z_{2} | = k$
$k = | z_{1} | | z_{2} | = 4$

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Standard XII Mathematics

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