The value of Arg `[i ln ((a-ib)/(a+ib))]`, where a and b are real numbers, is

Options

A. `0 or pi`
B. `pi/2`
C. not defined
D. None of these

Answer

To find the value of \( \text{Arg} \left[ i \ln \left( \frac{a - ib}{a + ib} \right) \right] \), where \( a \) and \( b \) are real numbers, we can follow these steps:

### Step 1: Rewrite the complex fraction
Let \( z = a + ib \). Then, the conjugate \( \overline{z} = a - ib \). We can express the fraction as:
\[ \frac{a - ib}{a + ib} = \frac{\overline{z}}{z} \]

### Step 2: Calculate the modulus and argument
The modulus of \( z \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
Thus, the modulus of \( \overline{z} \) is the same:
\[ |\overline{z}| = \sqrt{a^2 + b^2} \]
The argument \( \theta \) of \( z \) is:
\[ \theta = \tan^{-1} \left( \frac{b}{a} \right) \]
The argument of \( \overline{z} \) is:
\[ -\theta = -\tan^{-1} \left( \frac{b}{a} \right) \]

### Step 3: Find the argument of the fraction
Using the properties of arguments:
\[ \text{Arg} \left( \frac{\overline{z}}{z} \right) = \text{Arg}(\overline{z}) - \text{Arg}(z) = -\theta - \theta = -2\theta \]

### Step 4: Substitute into the logarithm
Now, we can express the logarithm:
\[ \ln \left( \frac{a - ib}{a + ib} \right) = \ln \left( |z| \right) + i \text{Arg} \left( \frac{a - ib}{a + ib} \right) \]
Substituting the argument we found:
\[ \ln \left( \frac{a - ib}{a + ib} \right) = \ln \left( \sqrt{a^2 + b^2} \right) + i(-2\theta) \]

### Step 5: Multiply by \( i \)
Now, we multiply by \( i \):
\[ i \ln \left( \frac{a - ib}{a + ib} \right) = i \ln \left( \sqrt{a^2 + b^2} \right) + i^2(-2\theta) \]
Since \( i^2 = -1 \), this simplifies to:
\[ i \ln \left( \sqrt{a^2 + b^2} \right) - 2\theta \]

### Step 6: Find the argument
The argument of \( i \ln \left( \sqrt{a^2 + b^2} \right) - 2\theta \) is determined by the real and imaginary parts. Since \( \ln \left( \sqrt{a^2 + b^2} \right) \) is a real number, the argument will depend on the sign of the imaginary part:
\[ \text{Arg} \left( i \ln \left( \sqrt{a^2 + b^2} \right) - 2\theta \right) \]
If \( -2\theta \) is real and \( \ln \left( \sqrt{a^2 + b^2} \right) \) is positive, the overall argument will be zero.

### Final Result
Thus, we conclude:
\[ \text{Arg} \left[ i \ln \left( \frac{a - ib}{a + ib} \right) \right] = 0 \]