Solve this following


Let a complex number be $w=1-\sqrt{3} i$. Let

another complex number $\mathrm{z}$ be such that $|\mathrm{zw}|=1$

and $\arg (\mathrm{z})-\arg (\mathrm{w})=\frac{\pi}{2}$. Then the area of the

triangle with vertices origin, $\mathrm{z}$ and $\mathrm{w}$ is equal to :


  1. 4

  2. $\frac{1}{2}$

  3. $\frac{1}{4}$

  4. 2

Correct Option: , 2


$\mathrm{w}=1-\sqrt{3} \cdot i \Rightarrow|\mathrm{w}|=2$

Now, $|z|=\frac{1}{|w|} \Rightarrow|z|=\frac{1}{2}$

and $\operatorname{amp}(\mathrm{z})=\frac{\pi}{2}+\operatorname{amp}(\mathrm{w})$

$\Rightarrow$ Area of triangle $=\frac{1}{2} . \mathrm{OP} . \mathrm{OQ}$

$=\frac{1}{2} \cdot 2 \cdot \frac{1}{2}=\frac{1}{2}$


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