Question

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

• A. 4
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. 2

Answer 1

Answer: (B)

$w=1-\sqrt{3}.i\Rightarrow |w|=2$
Now, $|z|=\frac{1}{|w|}\Rightarrow |z|=\frac{1}{2}$
and $\mathrm{amp}(z)=\frac{\pi }{2}+\mathrm{amp}(w)$

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