Question

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

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

Answer 1

The correct option is A $\frac{1}{2}$

$w=1-\sqrt{3}i\Rightarrow |w|=2\Rightarrow |zw|=1\Rightarrow |z|=\frac{1}{|w|}=\frac{1}{2}\arg \left(z\right)-\arg \left(w\right)=\frac{\pi }{2}$


Therefore, the area is
$\Delta =\frac{1}{2}\times \frac{1}{2}\cdot 2\therefore \Delta =\frac{1}{2}$