Suppose `arg (z) = - 5 pi//13`, then `arg((z + bar(z))/(1+z bar(z)))` is

Options

A. `- 5 pi//13`
B. `5 pi//13`
C. `pi`
D. 0

Answer

To solve the problem, we need to find the argument of the expression \(\frac{z + \bar{z}}{1 + z \bar{z}}\) given that \(\arg(z) = -\frac{5\pi}{13}\).

### Step-by-Step Solution:

1. **Understanding the Argument of \(z\)**:
Given \(\arg(z) = -\frac{5\pi}{13}\), we can express \(z\) in terms of its polar form:
\[ z = r(\cos(-\frac{5\pi}{13}) + i\sin(-\frac{5\pi}{13})) \]
where \(r\) is the modulus of \(z\).

2. **Finding \(z + \bar{z}\)**:
The conjugate of \(z\) is given by:
\[ \bar{z} = r(\cos(\frac{5\pi}{13}) + i\sin(\frac{5\pi}{13})) \]
Therefore, we have:
\[ z + \bar{z} = r(\cos(-\frac{5\pi}{13}) + i\sin(-\frac{5\pi}{13})) + r(\cos(\frac{5\pi}{13}) + i\sin(\frac{5\pi}{13})) \]
Simplifying this, we find:
\[ z + \bar{z} = r\left(2\cos\left(-\frac{5\pi}{13}\right)\right) = 2r\cos\left(-\frac{5\pi}{13}\right) \]
Since \(\cos(-\theta) = \cos(\theta)\), we have:
\[ z + \bar{z} = 2r\cos\left(\frac{5\pi}{13}\right) \]

3. **Finding \(1 + z \bar{z}\)**:
The product \(z \bar{z}\) is:
\[ z \bar{z} = r^2 \]
Therefore:
\[ 1 + z \bar{z} = 1 + r^2 \]

4. **Forming the Expression**:
Now we can substitute these results into the expression:
\[ \frac{z + \bar{z}}{1 + z \bar{z}} = \frac{2r\cos\left(\frac{5\pi}{13}\right)}{1 + r^2} \]

5. **Finding the Argument**:
The argument of a real number is 0 if it is positive, \(\pi\) if it is negative, and undefined if it is zero. Since \(r\) is a positive real number, \(1 + r^2\) is also positive. Therefore, the argument of the numerator \(2r\cos\left(\frac{5\pi}{13}\right)\) will depend on the sign of \(\cos\left(\frac{5\pi}{13}\right)\).

Since \(-\frac{5\pi}{13}\) is in the fourth quadrant, \(\cos\left(\frac{5\pi}{13}\right)\) is positive. Thus:
\[ \arg\left(\frac{z + \bar{z}}{1 + z \bar{z}}\right) = \arg(2r\cos\left(\frac{5\pi}{13}\right)) - \arg(1 + r^2) = 0 \]

6. **Final Result**:
Therefore, the argument of the expression is:
\[ \arg\left(\frac{z + \bar{z}}{1 + z \bar{z}}\right) = 0 \]