Let a complex number be w = 1 - sqrt{3} i. Le | vedclass

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(B) Given $w = 1 - \sqrt{3} i$,the modulus is $|w| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$.Given $|zw| = 1$,we have $|z| |w| = 1$,so $|z| = \

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$\frac{1}{2}$

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(B) Given $w = 1 - \sqrt{3} i$,the modulus is $|w| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$.Given $|zw| = 1$,we have $|z| |w| = 1$,so $|z| = \frac{1}{|w|} = \frac{1}{2}$.Given $\arg(z) - \arg(w) = \frac{\pi}{2}$,the angle between the vectors representing $z$ and $w$ at the origin is $\frac{\pi}{2}$.The area of a triangle with two sides of lengths $a$ and $b$ and an included angle $	heta$ is given by $\frac{1}{2} ab \sin(	heta)$.Here,the sides are $|z|$ and $|w|$,and the angle is $\frac{\pi}{2}$.Area $= \frac{1}{2} |z| |w| \sin\left(\frac{\pi}{2}ight) = \frac{1}{2} 	imes \frac{1}{2} 	imes 2 	imes 1 = \frac{1}{2}$.

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 at the origin,$z$,and $w$ is equal to ........ .

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