Question:
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 :
Correct Option: , 2
Solution:
$\mathrm{w}=1-\sqrt{3} \cdot i \Rightarrow|\mathrm{w}|=2$
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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