If |z| = 3, the area of the triangle whose sides are `z, omega z and z + omega z` (where `omega` is a complex cube root of unity) is

Options

A. `9 sqrt(3)//4`
B. `3 sqrt(3)//2`
C. `5//2`
D. `8 sqrt(3)//3`

Answer

To find the area of the triangle whose sides are \( z \), \( \omega z \), and \( z + \omega z \) where \( \omega \) is a complex cube root of unity and \( |z| = 3 \), we can follow these steps:

### Step 1: Identify the sides of the triangle
Given:
- \( |z| = 3 \)
- \( \omega \) is a complex cube root of unity, which satisfies \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

The sides of the triangle are:
- Side \( a = |z| = 3 \)
- Side \( b = |\omega z| = |\omega| \cdot |z| = 1 \cdot 3 = 3 \)
- Side \( c = |z + \omega z| = |z(1 + \omega)| \)

### Step 2: Calculate \( |1 + \omega| \)
Using the property of cube roots of unity:
- \( \omega = e^{2\pi i / 3} \)
- \( 1 + \omega = 1 + e^{2\pi i / 3} \)

To find the modulus:
\[ |1 + \omega| = |1 + e^{2\pi i / 3}| = |1 + (-\frac{1}{2} + i\frac{\sqrt{3}}{2})| = |1 - \frac{1}{2} + i\frac{\sqrt{3}}{2}| \]
\[ = | \frac{1}{2} + i\frac{\sqrt{3}}{2} | = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

Thus,
\[ |z + \omega z| = |z| \cdot |1 + \omega| = 3 \cdot 1 = 3 \]

### Step 3: Determine the sides of the triangle
Now we have:
- Side \( a = 3 \)
- Side \( b = 3 \)
- Side \( c = 3 \)

### Step 4: Area of the triangle
Since all sides are equal, the triangle is equilateral. The area \( A \) of an equilateral triangle with side length \( s \) is given by:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Substituting \( s = 3 \):
\[ A = \frac{\sqrt{3}}{4} \cdot 3^2 = \frac{\sqrt{3}}{4} \cdot 9 = \frac{9\sqrt{3}}{4} \]

### Final Answer
The area of the triangle is \( \frac{9\sqrt{3}}{4} \).

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