If ` arg ""( z^(1//3)) = (1)/(2) arg"" ( z^(2) + bar(z) z^(1//3))` then | z| =

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the problem, we start with the given equation:

\[ \arg(z^{1/3}) = \frac{1}{2} \arg(z^2 + \overline{z} z^{1/3}) \]

1. **Express the arguments**:
Let \( z = re^{i\theta} \), where \( r = |z| \) and \( \theta = \arg(z) \). Then we have:
\[ z^{1/3} = r^{1/3} e^{i\theta/3} \]
and
\[ z^2 = r^2 e^{2i\theta}, \quad \overline{z} = re^{-i\theta} \]
Therefore,
\[ \overline{z} z^{1/3} = r^{4/3} e^{i(\theta/3 - \theta)} = r^{4/3} e^{-2i\theta/3} \]

2. **Combine the terms**:
Now, we need to evaluate \( z^2 + \overline{z} z^{1/3} \):
\[ z^2 + \overline{z} z^{1/3} = r^2 e^{2i\theta} + r^{4/3} e^{-2i\theta/3} \]

3. **Find the argument of the sum**:
The argument of a sum of complex numbers can be tricky, but we can express it as:
\[ \arg(z^2 + \overline{z} z^{1/3}) = \arg(r^2 e^{2i\theta} + r^{4/3} e^{-2i\theta/3}) \]

4. **Set up the equation**:
From the original equation, we have:
\[ \frac{1}{3} \theta = \frac{1}{2} \arg(r^2 e^{2i\theta} + r^{4/3} e^{-2i\theta/3}) \]

5. **Simplify the equation**:
Rearranging gives:
\[ 2\theta = 3 \arg(r^2 e^{2i\theta} + r^{4/3} e^{-2i\theta/3}) \]

6. **Use properties of arguments**:
For the argument to be real, the expression inside the argument must be real. Hence, we can set up the condition:
\[ r^{2} + r^{4/3} \cos(2\theta - \frac{2\theta}{3}) = 0 \]

7. **Solve for \( r \)**:
We simplify the equation:
\[ r^{2} + r^{4/3} \cos(\frac{4\theta}{3}) = 0 \]
This implies:
\[ r^{2} = -r^{4/3} \cos(\frac{4\theta}{3}) \]
Since \( r \) must be non-negative, we can conclude that \( \cos(\frac{4\theta}{3}) \) must be zero.

8. **Find the modulus**:
The only way for this equation to hold true is if:
\[ |z| = 1 \]

Thus, the final answer is:

\[ |z| = 1 \]