Let a complex number be ω=1−3i. Let another complex number z be such that |zω|=1 and arg⁡(z)−arg⁡(ω)=π2 . Then the area of the triangle with vertices origin, z and ω is equal to

ω=1−3⋅i⇒|ω|=2Now |z|=1|ω|⇒|z|=12and amp⁡(z)=π2+amp⁡(ω)⇒ Area of triangle =12⋅OP⋅OQ=12⋅2⋅12=12

Let a complex number be ω=1−3i. Let another complex number z be such that |zω|=1 and arg⁡(z)−arg⁡(ω)=π2 . Then the area of the triangle with vertices origin, z and ω is equal to

ω=1−3⋅i⇒|ω|=2Now |z|=1|ω|⇒|z|=12and amp⁡(z)=π2+amp⁡(ω)⇒ Area of triangle =12⋅OP⋅OQ=12⋅2⋅12=12

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Let a complex number be $ω = 1 - \sqrt{3} i$ . Let another complex number z be such that $| zω | = 1$ and $\arg (z) - \arg (ω) = \frac{π}{2}$ . Then the area of the triangle with vertices origin, z and $ω$ is equal to

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Detailed Solution

$ω = 1 - \sqrt{3} \cdot i \Rightarrow | ω | = 2$
Now $| z | = \frac{1}{| ω |} \Rightarrow | z | = \frac{1}{2}$
and $amp (z) = \frac{π}{2} + amp (ω)$
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$\begin{array}{l} \Rightarrow Area of triangle = \frac{1}{2} \cdot OP \cdot OQ \\ = \frac{1}{2} \cdot 2 \cdot \frac{1}{2} = \frac{1}{2} \end{array}$