If the area of the triangle on the complex plane formed by complex numbers ` z, omega z and z + omega z ` is `4 sqrt(3)` square units, then | z| is

Options

A. 4
B. 2
C. 6
D. 3

Answer

To solve the problem, we need to find the modulus of the complex number \( z \) given that the area of the triangle formed by the complex numbers \( z \), \( \omega z \), and \( z + \omega z \) is \( 4\sqrt{3} \) square units.

### Step-by-Step Solution:

1. **Identify the Points**:
Let \( z \) be represented as a complex number. The point \( \omega z \) represents a rotation of \( z \) by an angle of \( \frac{\pi}{3} \) (since \( \omega = e^{i\frac{\pi}{3}} \)). The point \( z + \omega z \) is simply the vector sum of \( z \) and \( \omega z \).

2. **Understanding the Area of the Triangle**:
The area \( A \) of a triangle formed by three points \( A \), \( B \), and \( C \) in the complex plane can be calculated using the formula:
\[ A = \frac{1}{2} \left| \text{Im} \left( (B - A) \overline{(C - A)} \right) \right| \]
However, in this case, we can use the formula for the area of a triangle given by two sides and the included angle:
\[ A = \frac{1}{2} \cdot |z| \cdot |\omega z| \cdot \sin(\theta) \]
where \( \theta \) is the angle between the two sides.

3. **Calculate the Moduli**:
Since \( |\omega z| = |z| \) (as \( |\omega| = 1 \)), we can denote \( |z| = r \). Thus, we have:
\[ A = \frac{1}{2} \cdot r \cdot r \cdot \sin\left(\frac{\pi}{3}\right) \]
The sine of \( \frac{\pi}{3} \) is \( \frac{\sqrt{3}}{2} \).

4. **Substituting Values**:
Substituting the values into the area formula gives:
\[ A = \frac{1}{2} \cdot r^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^2 \]

5. **Setting Up the Equation**:
We know from the problem statement that the area \( A = 4\sqrt{3} \). Therefore, we can set up the equation:
\[ \frac{\sqrt{3}}{4} r^2 = 4\sqrt{3} \]

6. **Solving for \( r^2 \)**:
To solve for \( r^2 \), we multiply both sides by \( 4 \):
\[ \sqrt{3} r^2 = 16\sqrt{3} \]
Dividing both sides by \( \sqrt{3} \):
\[ r^2 = 16 \]

7. **Finding \( |z| \)**:
Taking the square root of both sides gives:
\[ |z| = 4 \]

### Final Answer:
Thus, the modulus of \( z \) is \( |z| = 4 \).

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