If `|z| = 1, z ne 1`, then value of arg `((1)/(1-z))` cannot exceed

Options

A. `pi//2`
B. `pi`
C. `3pi//2`
D. `2pi`

Answer

To solve the problem, we need to find the maximum value of the argument of the complex number \(\frac{1}{1-z}\) given that \(|z| = 1\) and \(z \neq 1\).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Since \(|z| = 1\), we can express \(z\) in terms of its polar form:
\[ z = e^{i\theta} \]
where \(\theta\) is the argument of \(z\).

2. **Substituting \(z\)**:
We substitute \(z\) into the expression \(\frac{1}{1-z}\):
\[ \frac{1}{1-z} = \frac{1}{1 - e^{i\theta}} \]

3. **Finding the Denominator**:
To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{1}{1 - e^{i\theta}} \cdot \frac{1 - e^{-i\theta}}{1 - e^{-i\theta}} = \frac{1 - e^{-i\theta}}{(1 - e^{i\theta})(1 - e^{-i\theta})} \]

4. **Calculating the Denominator**:
The denominator simplifies as follows:
\[ (1 - e^{i\theta})(1 - e^{-i\theta}) = 1 - (e^{i\theta} + e^{-i\theta}) + 1 = 2 - 2\cos(\theta) = 2(1 - \cos(\theta)) \]

5. **Expressing the Result**:
Thus, we have:
\[ \frac{1}{1 - z} = \frac{1 - e^{-i\theta}}{2(1 - \cos(\theta))} \]

6. **Finding the Argument**:
The argument of \(\frac{1 - e^{-i\theta}}{2(1 - \cos(\theta))}\) can be expressed as:
\[ \arg\left(1 - e^{-i\theta}\right) - \arg\left(2(1 - \cos(\theta))\right) \]
Since \(2(1 - \cos(\theta))\) is a positive real number, its argument is \(0\).

7. **Calculating \(\arg(1 - e^{-i\theta})\)**:
We need to find \(\arg(1 - e^{-i\theta})\):
\[ 1 - e^{-i\theta} = 1 - (\cos(\theta) - i\sin(\theta)) = (1 - \cos(\theta)) + i\sin(\theta) \]
The argument is given by:
\[ \tan^{-1}\left(\frac{\sin(\theta)}{1 - \cos(\theta)}\right) \]

8. **Using Trigonometric Identities**:
We can use the identity \(1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2})\) to rewrite:
\[ \tan^{-1}\left(\frac{\sin(\theta)}{2\sin^2(\frac{\theta}{2})}\right) = \tan^{-1}\left(\frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{2\sin^2(\frac{\theta}{2})}\right) = \tan^{-1}\left(\cot(\frac{\theta}{2})\right) \]

9. **Finding Maximum Value**:
The maximum value of \(\tan^{-1}(\cot(\frac{\theta}{2}))\) approaches \(\frac{\pi}{2}\) as \(\theta\) approaches \(0\) or \(2\pi\), but since \(z \neq 1\), \(\theta\) cannot be exactly \(0\).

10. **Conclusion**:
Therefore, the maximum value of \(\arg\left(\frac{1}{1-z}\right)\) cannot exceed \(\frac{\pi}{2}\).

### Final Answer:
The value of \(\arg\left(\frac{1}{1-z}\right)\) cannot exceed \(\frac{\pi}{2}\).