Question:
Let a complex number be $w=1-\sqrt{3} i$. Let another complex number $\mathrm{z}$ be such that $|z w|=1$ and $\arg (z)-\arg (w)=\frac{\pi}{2}$. Then the area of the triangle with vertices origin, $\mathrm{z}$ and $\mathrm{w}$ is equal to:
Correct Option: , 2
Solution:
$\mathrm{w}=1-\sqrt{3} \cdot i \Rightarrow \mid \mathrm{wl}=2$
Now, $|z|=\frac{1}{|w|} \Rightarrow|z|=\frac{1}{2}$
and $\operatorname{amp}(\mathrm{z})=\frac{\pi}{2}+\mathrm{amp}(\mathrm{w})$
$\Rightarrow$ Area of triangle $=\frac{1}{2} \cdot$ OP. OQ
$=\frac{1}{2} \cdot 2 \cdot \frac{1}{2}=\frac{1}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.