For a complex number Z, if arg `Z=(pi)/(4)` and `|Z+(1)/(Z)|=4`, then the value of `||Z|-(1)/(|Z|)|` is equal to

Options

A. `sqrt(14)`
B. `sqrt(18)`
C. 4
D. `sqrt(12)`

Answer

To solve the problem, we need to find the value of \( ||Z| - \frac{1}{|Z|} | \) given that \( \arg Z = \frac{\pi}{4} \) and \( |Z + \frac{1}{Z}| = 4 \).

### Step 1: Express \( Z \) in terms of its modulus and argument
Since \( \arg Z = \frac{\pi}{4} \), we can express \( Z \) in polar form:
\[ Z = r \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = r \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) \]
where \( r = |Z| \).

### Step 2: Calculate \( \frac{1}{Z} \)
The reciprocal of \( Z \) is given by:
\[ \frac{1}{Z} = \frac{1}{r} \left( \cos \left(-\frac{\pi}{4}\right) + i \sin \left(-\frac{\pi}{4}\right) \right) = \frac{1}{r} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \]

### Step 3: Calculate \( Z + \frac{1}{Z} \)
Now we can compute \( Z + \frac{1}{Z} \):
\[ Z + \frac{1}{Z} = r \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) + \frac{1}{r} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \]
Combining the real and imaginary parts:
\[ = \left( r + \frac{1}{r} \right) \frac{1}{\sqrt{2}} + i \left( r - \frac{1}{r} \right) \frac{1}{\sqrt{2}} \]

### Step 4: Calculate the modulus
The modulus of \( Z + \frac{1}{Z} \) is given by:
\[ |Z + \frac{1}{Z}| = \sqrt{ \left( r + \frac{1}{r} \right)^2 \frac{1}{2} + \left( r - \frac{1}{r} \right)^2 \frac{1}{2} } \]
This simplifies to:
\[ = \sqrt{ \frac{1}{2} \left( (r + \frac{1}{r})^2 + (r - \frac{1}{r})^2 \right) } \]

### Step 5: Expand and simplify
Expanding the squares:
\[ (r + \frac{1}{r})^2 = r^2 + 2 + \frac{1}{r^2} \]
\[ (r - \frac{1}{r})^2 = r^2 - 2 + \frac{1}{r^2} \]
Adding these:
\[ (r + \frac{1}{r})^2 + (r - \frac{1}{r})^2 = 2r^2 + 2\frac{1}{r^2} \]
Thus,
\[ |Z + \frac{1}{Z}| = \sqrt{ \frac{1}{2} \left( 2r^2 + 2\frac{1}{r^2} \right) } = \sqrt{ r^2 + \frac{1}{r^2} } \]

### Step 6: Set the modulus equal to 4
Given that \( |Z + \frac{1}{Z}| = 4 \):
\[ \sqrt{ r^2 + \frac{1}{r^2} } = 4 \]
Squaring both sides:
\[ r^2 + \frac{1}{r^2} = 16 \]

### Step 7: Rearranging the equation
Let \( x = r \), then:
\[ x^2 + \frac{1}{x^2} = 16 \]
Multiplying through by \( x^2 \):
\[ x^4 - 16x^2 + 1 = 0 \]
Let \( y = x^2 \):
\[ y^2 - 16y + 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{16 \pm \sqrt{256 - 4}}{2} = \frac{16 \pm \sqrt{252}}{2} = 8 \pm \sqrt{63} \]

### Step 8: Find \( ||Z| - \frac{1}{|Z|}| \)
Now we need to find:
\[ ||r| - \frac{1}{|r|}| \]
Since \( r^2 = 8 + \sqrt{63} \) or \( r^2 = 8 - \sqrt{63} \), we can calculate \( |r| - \frac{1}{|r|} \) for both cases.

### Final Calculation
Let’s take \( r^2 = 8 + \sqrt{63} \):
\[ r = \sqrt{8 + \sqrt{63}}, \quad \frac{1}{r} = \frac{1}{\sqrt{8 + \sqrt{63}}} \]
Calculating \( ||r| - \frac{1}{|r|}| \):
\[ ||\sqrt{8 + \sqrt{63}} - \frac{1}{\sqrt{8 + \sqrt{63}}}| \]
This will yield a specific numerical value.

### Conclusion
The final answer is:
\[ ||Z| - \frac{1}{|Z|}| = \sqrt{14} \]