The value of `(1+isqrt(3))/(1-isqrt(3))^(6)+(1-isqrt(3))/(1+isqrt(3))^(6)` is

Options

A. 2
B. `-2`
C. 1/32
D. 0

Answer

To solve the expression \(\frac{1 + i\sqrt{3}}{(1 - i\sqrt{3})^6} + \frac{1 - i\sqrt{3}}{(1 + i\sqrt{3})^6}\), we can follow these steps:

### Step 1: Simplify the individual terms
Let's denote \( z_1 = 1 + i\sqrt{3} \) and \( z_2 = 1 - i\sqrt{3} \). We can rewrite the expression as:
\[ \frac{z_1}{z_2^6} + \frac{z_2}{z_1^6} \]

### Step 2: Calculate the modulus and argument of \( z_1 \) and \( z_2 \)
The modulus of \( z_1 \) is:
\[ |z_1| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
The argument of \( z_1 \) is:
\[ \theta_1 = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Thus, we can express \( z_1 \) in polar form:
\[ z_1 = 2 \text{cis}\left(\frac{\pi}{3}\right) \]

Similarly, for \( z_2 \):
\[ |z_2| = 2 \quad \text{and} \quad \theta_2 = -\frac{\pi}{3} \]
So,
\[ z_2 = 2 \text{cis}\left(-\frac{\pi}{3}\right) \]

### Step 3: Raise \( z_1 \) and \( z_2 \) to the 6th power
Using De Moivre's theorem:
\[ z_1^6 = (2 \text{cis}\left(\frac{\pi}{3}\right))^6 = 2^6 \text{cis}(2\pi) = 64 \cdot 1 = 64 \]
\[ z_2^6 = (2 \text{cis}\left(-\frac{\pi}{3}\right))^6 = 64 \cdot 1 = 64 \]

### Step 4: Substitute back into the expression
Now substituting back into the original expression:
\[ \frac{z_1}{z_2^6} + \frac{z_2}{z_1^6} = \frac{1 + i\sqrt{3}}{64} + \frac{1 - i\sqrt{3}}{64} \]

### Step 5: Combine the fractions
Combining the two fractions:
\[ = \frac{(1 + i\sqrt{3}) + (1 - i\sqrt{3})}{64} = \frac{2}{64} = \frac{1}{32} \]

### Final Answer
Thus, the value of the expression is:
\[ \frac{1}{32} \]