If `f(z) = (7-z)/( 1-z^(2))`, where `z =1+2i`, then `|f (z) |` is

Options

A. `(|z|)/(2)`
B. `|z|`
C. `2|z|`
D. None of these

Answer

To solve the problem, we need to find the modulus of the function \( f(z) = \frac{7 - z}{1 - z^2} \) where \( z = 1 + 2i \).

### Step-by-Step Solution

1. **Calculate \( z^2 \)**:
\[ z = 1 + 2i \]
\[ z^2 = (1 + 2i)^2 = 1^2 + 2 \cdot 1 \cdot 2i + (2i)^2 = 1 + 4i - 4 = -3 + 4i \]

2. **Substitute \( z \) and \( z^2 \) into \( f(z) \)**:
\[ f(z) = \frac{7 - (1 + 2i)}{1 - (-3 + 4i)} \]
Simplifying the numerator:
\[ 7 - (1 + 2i) = 6 - 2i \]
Simplifying the denominator:
\[ 1 - (-3 + 4i) = 1 + 3 - 4i = 4 - 4i \]
Thus,
\[ f(z) = \frac{6 - 2i}{4 - 4i} \]

3. **Rationalize the denominator**:
Multiply the numerator and denominator by the conjugate of the denominator:
\[ f(z) = \frac{(6 - 2i)(4 + 4i)}{(4 - 4i)(4 + 4i)} \]
Calculating the denominator:
\[ (4 - 4i)(4 + 4i) = 16 + 16 = 32 \]
Calculating the numerator:
\[ (6 - 2i)(4 + 4i) = 24 + 24i - 8i - 8 = 16 + 16i \]
Thus,
\[ f(z) = \frac{16 + 16i}{32} = \frac{1 + i}{2} \]

4. **Find the modulus \( |f(z)| \)**:
The modulus of a complex number \( a + bi \) is given by:
\[ |a + bi| = \sqrt{a^2 + b^2} \]
Here, \( a = \frac{1}{2} \) and \( b = \frac{1}{2} \):
\[ |f(z)| = \left| \frac{1}{2} + \frac{1}{2}i \right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

5. **Express \( |f(z)| \) in terms of \( |z| \)**:
We know that \( |z| = |1 + 2i| = \sqrt{1^2 + 2^2} = \sqrt{5} \).
We can express \( |f(z)| \) as:
\[ |f(z)| = \frac{|z|}{2} \]

### Final Answer
Thus, the modulus \( |f(z)| \) is:
\[ |f(z)| = \frac{\sqrt{5}}{2} \]