Let a complex number be w = 1 - √3i. Let another complex number z be such that |zw| = 1 and arg(z) -

Q: Wait, do I need to find the exact coordinates of $z$ to solve this? That seems like a lot of algebra.

Absolutely not! That is the trap. JEE Advanced loves to test if you can see the geometry behind the algebra. By treating $z$ and $w$ as vectors, we bypass the need for coordinates entirely. Geometry is your best friend here.

Q: Does the quadrant of $w$ affect the area?

Great question! The area of a triangle depends only on the lengths of the sides and the angle between them. Since the angle is fixed at $\frac{\pi}{2}$ and the magnitudes are fixed, the area remains invariant regardless of which quadrant $w$ sits in.

Q: Why is the area formula $\frac{1}{2} |z| |w| \sin(\theta)$ valid here?

Think of the triangle formed by the origin, $z$, and $w$. This is a classic triangle with two sides of lengths $|z|$ and $|w|$ meeting at the origin with an included angle $\theta$. This is the fundamental definition of the area of a triangle in vector geometry, and it holds perfectly in the Argand plane.

Question

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 origin, $z$ and $w$ is equal to:

Options:

(A) 4

(B) $\frac{1}{2}$

(C) $\frac{1}{4}$

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving an equation; we are painting a picture in the Argand plane.

When you see a complex number like $w = 1 - \sqrt{3}i$, do not just see a static value. See a vector. See a directed line segment stretching from the origin $O(0,0)$ to the point $(1, -\sqrt{3})$. This is the anchor of our problem.

### Decoding the Vector $w$

First, let us understand the magnitude of our vector $w$. We calculate the modulus:

$$|w| = \sqrt{1^2 + (-\sqrt{3})^2}$$

This simplifies to $\sqrt{1 + 3} = \sqrt{4} = 2$. So, we have a vector of length $2$ units.

It sits in the fourth quadrant, but for the area of our triangle, the specific quadrant is merely a detail. The magnitude is what matters.

### The Modulus Constraint

Now, look at the condition $|zw| = 1$. This is where the magic of complex algebra shines.

We know the property $|zw| = |z| \cdot |w|$. Substituting our known value, we get:

$$|z| \cdot 2 = 1 \Rightarrow |z| = \frac{1}{2}$$

We have just discovered that $z$ is a much shorter vector, only half a unit long.

### The Geometric Aha! Moment

The problem states $\arg(z) - \arg(w) = \frac{\pi}{2}$. This is the most critical piece of the puzzle.

In the language of vectors, the difference in arguments is simply the angle between them. This means the angle between the vector $Oz$ and the vector $Ow$ is exactly $\frac{\pi}{2}$, or $90^\circ$.

We are looking at a right-angled triangle!

### The Final Calculation

We have a triangle with two sides of lengths $|z| = \frac{1}{2}$ and $|w| = 2$, with an included angle of $\frac{\pi}{2}$. The area of any triangle with sides $a$ and $b$ and included angle $	heta$ is given by:

$$Area = \frac{1}{2} |z| |w| \sin(	heta)$$

Substituting our values, we get:

$$Area = \frac{1}{2} \cdot \left(\frac{1}{2}ight) \cdot (2) \cdot \sin\left(\frac{\pi}{2}ight)$$

Since $\sin(\frac{\pi}{2}) = 1$, the calculation becomes:

$$Area = \frac{1}{2} \cdot \left(\frac{1}{2}ight) \cdot (2) \cdot (1) = \frac{1}{2}$$

The $2$ and the $\frac{1}{2}$ cancel out beautifully, leaving us with an area of **$\frac{1}{2}$**. See how elegant that is? By trusting the geometry, we avoided the messy algebra of finding the coordinates of $z$ entirely.

Animated Solution for Mathematics - Complex Numbers: Let a complex number be $w = 1 - 3 i$ . Let another complex number $z$ be such that $∣ z w ∣ = 1$ 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:

Select Answer:

Visualized Solution

The Argand Plane

Plotting Complex Number $w$

Modulus of $w$

The Modulus Condition

Modulus of $z$

The Argument Condition

Triangle $O z w$

Area of Triangle Formula

Substituting Values

Calculating the Area

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Decoding the Vector $w$

The Modulus Constraint

The Geometric Aha! Moment

The Final Calculation

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Select Answer:

Visualized Solution

The Argand Plane

Plotting Complex Number .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

Modulus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

The Modulus Condition

Modulus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

The Argument Condition

Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ozw

Area of Triangle Formula

Substituting Values

Calculating the Area

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Decoding the Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

The Modulus Constraint

The Geometric Aha! Moment

The Final Calculation

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