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

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Complex Numbers

Let a complex number be

w = 1 - 3 i

. Let another complex number z be such that

∣ zw ∣ = 1

and

ar g (z) - ar g (w) = \frac{π}{2}

. Then the area of the triangle with vertices origin,

z

and

w

is equal to :

Answer: (C)

From

w = 1 - 3 i

∣ w ∣ = 1 + 3 = 2

and

ar g w = - \frac{π}{3}

. The condition

∣ z w ∣ = 1

gives

∣ z ∣∣ w ∣ = 1

, so

∣ z ∣ = \frac{1}{2}

, and the angle between the segments joining the origin to

z

and to

w

ar g z - ar g w = \frac{π}{2}

. Those two segments are the sides of the triangle meeting at the origin, so

Area = \frac{1}{2} ∣ z ∣∣ w ∣ sin \frac{π}{2} = \frac{1}{2} \cdot \frac{1}{2} \cdot 2 \cdot 1 = \frac{1}{2}

The area of the triangle is

\frac{1}{2}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option