Let a complex number be w=1-√3 i.

Question

Let a complex number be $w=1-\sqrt3 i$. Let another complex number $z$ be such that $|zw|=1$ and $\arg(z)-\arg(w)=\dfrac\pi2$. Then the area of the triangle with vertices origin, $z$ and $w$, is equal to

Accepted Answer

$|w|=\sqrt{1+3}=2$, and $|zw|=1 \implies |z|=\dfrac12$. Since the angle between $z$ and $w$ is $\dfrac\pi2$,

$$\text{Area}=\frac12|z||w|\sin\frac\pi2=\frac12\cdot\frac12\cdot2=\frac12.$$

Solutions · 1