If z1- z2 are two complex numbers such that -z1-z2-2 and z1 - z2-0- then z1z2-

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 XIII

Mathematics

Argument of a Complex Number

If z1, z2 are...

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 $θ = arg (z_{1})$
$z_{2} = | z_{2} | (cos ϕ + i sin ϕ)$
Where $ϕ = arg (z_{2})$
It is given that $arg z_{2} = - arg z_{1} \Rightarrow ϕ = - θ$
Also,
$| z_{1} | = | z_{2} | z_{2} = | z_{1} | [cos (- θ) + i sin (- θ)] \Rightarrow z_{2} = | z_{1} | (cos θ - i sin θ) \Rightarrow z_{2} =_{1} \Rightarrow z_{1} =_{2} \Rightarrow z_{1} z_{2} =_{2} z_{2} = | z_{2} |^{2} = 4$

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

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Argument of a Complex Number

Standard XIII Mathematics

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If z1,z2 are two complex numbers such that |z1|=|z2|=2 and argz1+argz2=0, then z1z2=

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 $z_{1} z_{2} =$