$\text{ Let a complex number be }w=1-\sqrt{3}i\text{ . Let another complex number }z\text{ be such that }$$\left|zw\right|=1\text{ and }$

$\arg \left(z\right)-\arg \left(w\right)=\frac{\pi }{2}.\text{ Then the area of the triangle with vertices origin, }z\text{ and }w\text{ is equal }$

$\begin{array}{c}\text{Given w}=1-\sqrt{3}i\\ \left|w\right|=\sqrt{1+3}=\sqrt{4}\end{array}$

$\begin{array}{c}\left|zw\right|=1\\ \Rightarrow \left|z\right|.\left|w\right|=1\\ \Rightarrow \left|z\right|=\frac{1}{\left|w\right|}\\ =\frac{1}{2}\end{array}$

$\begin{array}{l}\arg \left(w\right)=-{\tan }^{-1}\left(\sqrt{3}\right)\\ =\frac{-\pi }{3}\end{array}$

$\begin{array}{l}\text{Given arg}\left(z\right)=\frac{\pi }{2}+\arg \left(w\right)\\ =\frac{\pi }{2}-\frac{\pi }{3}\\ =\frac{\pi }{6}\end{array}$

$z=\frac{1}{4}\left(\sqrt{3}+i\right)$

$\begin{array}{l}\text{The area of the triangle formed by the points O, z, w is }\\ \frac{1}{2}\left|x_1{y}_2-x_2{y}_1\right|=\frac{1}{2}\left|\frac{3}{4}+\frac{1}{4}\right|=\frac{1}{2}\left(1\right)=\frac{1}{2}\text{\hspace{0.17em}square units}\end{array}$