Let z be a complex number satisfying `1/2 le |z| le 4` , then sum of greatest and least values of `|z+1/z|` is :

Options

A. `65/4`
B. `65/16`
C. `17/4`
D. 17

Answer

To solve the problem, we need to find the sum of the greatest and least values of \( |z + \frac{1}{z}| \) given that \( \frac{1}{2} \leq |z| \leq 4 \).

### Step-by-Step Solution:

1. **Let \( z = r e^{i\theta} \)**:
Since \( z \) is a complex number, we can express it in polar form where \( r = |z| \) and \( \theta \) is the argument of \( z \).

2. **Express \( z + \frac{1}{z} \)**:
\[ z + \frac{1}{z} = r e^{i\theta} + \frac{1}{r} e^{-i\theta} \]
This can be rewritten as:
\[ z + \frac{1}{z} = r \cos \theta + i r \sin \theta + \frac{1}{r} \cos \theta - i \frac{1}{r} \sin \theta \]
Combining real and imaginary parts:
\[ = \left( r + \frac{1}{r} \right) \cos \theta + i \left( r - \frac{1}{r} \right) \sin \theta \]

3. **Find the modulus**:
The modulus of \( z + \frac{1}{z} \) is given by:
\[ |z + \frac{1}{z}| = \sqrt{\left( r + \frac{1}{r} \right)^2 \cos^2 \theta + \left( r - \frac{1}{r} \right)^2 \sin^2 \theta} \]

4. **Simplify the expression**:
Expanding the squares:
\[ = \sqrt{\left( r^2 + 2 + \frac{1}{r^2} \right) \cos^2 \theta + \left( r^2 - 2 + \frac{1}{r^2} \right) \sin^2 \theta} \]
\[ = \sqrt{r^2 \cos^2 \theta + \frac{1}{r^2} \cos^2 \theta + r^2 \sin^2 \theta + \frac{1}{r^2} \sin^2 \theta + 2 \cos^2 \theta - 2 \sin^2 \theta} \]
\[ = \sqrt{r^2 + \frac{1}{r^2} + 2 \cos^2 \theta - 2 \sin^2 \theta} \]
\[ = \sqrt{r^2 + \frac{1}{r^2} + 2 \cos(2\theta)} \]

5. **Determine the maximum and minimum values**:
- The term \( \cos(2\theta) \) varies between -1 and 1.
- Therefore, the maximum value occurs when \( \cos(2\theta) = 1 \):
\[ |z + \frac{1}{z}|_{\text{max}} = \sqrt{r^2 + \frac{1}{r^2} + 2} = \sqrt{\left(r + \frac{1}{r}\right)^2} \]
Thus, \( |z + \frac{1}{z}|_{\text{max}} = r + \frac{1}{r} \).
- The minimum value occurs when \( \cos(2\theta) = -1 \):
\[ |z + \frac{1}{z}|_{\text{min}} = \sqrt{r^2 + \frac{1}{r^2} - 2} = \sqrt{\left(r - \frac{1}{r}\right)^2} \]
Thus, \( |z + \frac{1}{z}|_{\text{min}} = r - \frac{1}{r} \).

6. **Evaluate for the given bounds of \( r \)**:
- For \( r = 4 \):
\[ |z + \frac{1}{z}|_{\text{max}} = 4 + \frac{1}{4} = \frac{17}{4} \]
\[ |z + \frac{1}{z}|_{\text{min}} = 4 - \frac{1}{4} = \frac{15}{4} \]
- For \( r = \frac{1}{2} \):
\[ |z + \frac{1}{z}|_{\text{max}} = \frac{1}{2} + 2 = \frac{5}{2} \]
\[ |z + \frac{1}{z}|_{\text{min}} = \frac{1}{2} - 2 = -\frac{3}{2} \text{ (not valid since modulus cannot be negative)} \]

7. **Sum of greatest and least values**:
The greatest value is \( \frac{17}{4} \) and the least value is \( 0 \) (since \( |z + \frac{1}{z}| \) cannot be negative).
\[ \text{Sum} = \frac{17}{4} + 0 = \frac{17}{4} \]

### Final Answer:
The sum of the greatest and least values of \( |z + \frac{1}{z}| \) is \( \frac{17}{4} \).