The maximum distance from the origin of coordinates of the point z satisfying the equation `| z + (1)/(z)| = a ` is

Options

A. `(1)/(2) ( sqrt(a^(2) + 1) + a)`
B. `(1)/(2) ( sqrt(a ^(2) + 2) + a)`
C. `(1)/(2) ( sqrt(a^(2) + 4) + a)`
D. none of these

Answer

To solve the problem of finding the maximum distance from the origin of the point \( z \) satisfying the equation \( | z + \frac{1}{z}| = a \), we will follow these steps:

### Step-by-Step Solution:

1. **Substitute \( z \)**:
Let \( z = re^{i\theta} \), where \( r \) is the modulus (distance from the origin) and \( \theta \) is the argument (angle).

2. **Rewrite the equation**:
The equation becomes:
\[ |re^{i\theta} + \frac{1}{re^{i\theta}}| = a \]
This simplifies to:
\[ |re^{i\theta} + \frac{1}{r} e^{-i\theta}| = a \]

3. **Combine the terms**:
We can express this as:
\[ |r \cos \theta + i r \sin \theta + \frac{1}{r} \cos \theta - i \frac{1}{r} \sin \theta| = a \]
This simplifies to:
\[ |(r + \frac{1}{r}) \cos \theta + i (r - \frac{1}{r}) \sin \theta| = a \]

4. **Square both sides**:
Squaring both sides gives:
\[ (r + \frac{1}{r})^2 \cos^2 \theta + (r - \frac{1}{r})^2 \sin^2 \theta = a^2 \]

5. **Expand the squares**:
Expanding the left-hand side results in:
\[ (r^2 + 2 + \frac{1}{r^2}) \cos^2 \theta + (r^2 - 2 + \frac{1}{r^2}) \sin^2 \theta = a^2 \]

6. **Combine terms**:
This can be rearranged to:
\[ r^2(\cos^2 \theta + \sin^2 \theta) + \frac{1}{r^2}(\cos^2 \theta + \sin^2 \theta) + 2\cos^2 \theta - 2\sin^2 \theta = a^2 \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we have:
\[ r^2 + \frac{1}{r^2} + 2\cos(2\theta) = a^2 \]

7. **Maximize the expression**:
To maximize \( r \), we need to minimize \( \cos(2\theta) \). The minimum value of \( \cos(2\theta) \) is -1, which occurs when \( 2\theta = \pi \) or \( \theta = \frac{\pi}{2} \).

8. **Substituting the minimum value**:
Substituting \( \cos(2\theta) = -1 \) into the equation gives:
\[ r^2 + \frac{1}{r^2} - 2 = a^2 \]
Rearranging this leads to:
\[ r^2 + \frac{1}{r^2} = a^2 + 2 \]

9. **Let \( x = r + \frac{1}{r} \)**:
We know that:
\[ r^2 + \frac{1}{r^2} = x^2 - 2 \]
Thus:
\[ x^2 - 2 = a^2 + 2 \implies x^2 = a^2 + 4 \]

10. **Solve for \( r \)**:
Therefore:
\[ r + \frac{1}{r} = \sqrt{a^2 + 4} \]
Let \( r + \frac{1}{r} = x \). The quadratic equation is:
\[ r^2 - xr + 1 = 0 \]
Solving this gives:
\[ r = \frac{x \pm \sqrt{x^2 - 4}}{2} \]
The maximum value of \( r \) occurs when we take the positive root:
\[ r = \frac{\sqrt{a^2 + 4} + \sqrt{(\sqrt{a^2 + 4})^2 - 4}}{2} \]
Simplifying gives:
\[ r = \frac{\sqrt{a^2 + 4} + \sqrt{a^2}}{2} = \frac{\sqrt{a^2 + 4} + a}{2} \]

### Final Answer:
The maximum distance from the origin of the point \( z \) satisfying the equation \( | z + \frac{1}{z}| = a \) is:
\[ \frac{a + \sqrt{a^2 + 4}}{2} \]