Question

Let a complex number be w = 1 - √3 i. Let another complex number z be such that |z w| = 1 and arg(z) - arg(w) = π/2. Then the area of the triangle with vertices origin, z and w is equal to :

• A. 1/2
• B. 2
• C. 1/4
• D. 4

Hint:

The area of a triangle with vertices $0, z_1, z_2$ is $\frac{1}{2} |z_1| |z_2| \sin(\theta_2 - \theta_1)$.

Answer 1

The Correct Option is A

To find the area of the triangle with vertices at the origin, \(z\), and \(w\) where \(w = 1 - \sqrt{3}i\) and \(z\) is such that \(|zw| = 1\) and \(\text{arg}(z) - \text{arg}(w) = \pi/2\), we can use the properties of complex numbers.

1. Start with the given complex number \(w = 1 - \sqrt{3}i\).
2. The modulus of \(w\) is calculated as: |w| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.
3. The argument \(\text{arg}(w)\) is given by: \text{arg}(w) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}.
4. For the complex number \(z\), it is given that \(|zw| = 1\). Therefore, |z| = \frac{1}{|w|} = \frac{1}{2}.
5. The condition \(\text{arg}(z) - \text{arg}(w) = \pi/2\) implies: \text{arg}(z) = \text{arg}(w) + \frac{\pi}{2} = -\frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{6}.
6. Therefore, \(z\) in polar form is given by: z = \frac{1}{2} (\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})\. This simplifies to: z = \frac{\sqrt{3}}{4} + i \frac{1}{4}\.
7. To find the area of the triangle with vertices at the origin, \(z\), and \(w\), use the formula for the area of the triangle formed by the origin and two complex numbers. The area \(A\) is given by: A = \frac{1}{2} \times \left| \text{Imaginary part of} (z \cdot \overline{w})\right|.
8. Calculate \(\overline{w}\), the complex conjugate of \(w\), as: \overline{w} = 1 + \sqrt{3}i.
9. Compute \(z \cdot \overline{w}\): z \cdot \overline{w} = \left(\frac{\sqrt{3}}{4} + i \frac{1}{4}\right) \cdot (1 + \sqrt{3}i). Simplifying, this yields: \frac{\sqrt{3}}{4} + \frac{3}{4}i + i \frac{1}{4} - i^2 \frac{\sqrt{3}}{12} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{12} + i \left(\frac{3}{4} + \frac{1}{4}\right) = \frac{\sqrt{3}}{3} + i\.
10. The imaginary part is \(1\), so the area is: A = \frac{1}{2} \times 1 = \frac{1}{2}.

Thus, the area of the triangle is \frac{1}{2}.