If a complex number z satisfies `|z| = 1` and `arg(z-1) = (2pi)/(3)`, then (`omega` is complex imaginary number)

Options

A. `z^(2) + z` is purely imaginary number
B. `z = - omega^(2)`
C. `z = - omega`
D. `|z-1| = 1` then,

Explanation

To solve the problem step by step, we will follow the conditions given in the question.

### Step 1: Define the complex number
Let \( z = x + yi \), where \( x \) and \( y \) are real numbers.

### Step 2: Apply the modulus condition
From the condition \( |z| = 1 \), we have:
\[ |z| = \sqrt{x^2 + y^2} = 1 \]
Squaring both sides gives:
\[ x^2 + y^2 = 1 \quad \text{(1)} \]

### Step 3: Apply the argument condition
The second condition states \( \arg(z - 1) = \frac{2\pi}{3} \). This means:
\[ \arg(x - 1 + yi) = \frac{2\pi}{3} \]
From the definition of argument, we have:
\[ \tan\left(\frac{2\pi}{3}\right) = \frac{y}{x - 1} \]
Calculating \( \tan\left(\frac{2\pi}{3}\right) \):
\[ \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3} \]
Thus, we can write:
\[ -\sqrt{3} = \frac{y}{x - 1} \]
Rearranging gives:
\[ y = -\sqrt{3}(x - 1) \quad \text{(2)} \]

### Step 4: Substitute \( y \) into the modulus equation
Substituting equation (2) into equation (1):
\[ x^2 + (-\sqrt{3}(x - 1))^2 = 1 \]
Expanding this:
\[ x^2 + 3(x - 1)^2 = 1 \]
\[ x^2 + 3(x^2 - 2x + 1) = 1 \]
\[ x^2 + 3x^2 - 6x + 3 = 1 \]
Combining like terms:
\[ 4x^2 - 6x + 2 = 0 \]

### Step 5: Simplify the quadratic equation
Dividing the entire equation by 2:
\[ 2x^2 - 3x + 1 = 0 \]

### Step 6: Factor the quadratic equation
Factoring gives:
\[ (2x - 1)(x - 1) = 0 \]
Thus, the roots are:
\[ x = \frac{1}{2} \quad \text{and} \quad x = 1 \]
Since \( x = 1 \) would make \( z - 1 \) indeterminate, we take:
\[ x = \frac{1}{2} \]

### Step 7: Find \( y \)
Substituting \( x = \frac{1}{2} \) back into equation (2):
\[ y = -\sqrt{3}\left(\frac{1}{2} - 1\right) = -\sqrt{3}\left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2} \]

### Step 8: Write the complex number \( z \)
Thus, we have:
\[ z = \frac{1}{2} + \frac{\sqrt{3}}{2} i \]

### Step 9: Relate \( z \) to \( \omega \)
We know that:
\[ \omega = e^{\frac{2\pi i}{3}} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \]
Thus, \( z \) can be expressed in terms of \( \omega \):
\[ z = -\omega^2 \]

### Step 10: Verify the options
1. **Option A**: Correct, since \( z = -\omega^2 \) is valid.
2. **Option B**: Correct, as \( z^2 = \omega \).
3. **Option C**: Incorrect, as \( z^2 + z = \omega - \omega^2 \) is purely imaginary.
4. **Option D**: Correct, since \( |z - 1| = 1 \).