The roots of the equation `(1+isqrt(3))^(x)-2^(x)=0` form

Options

A. an A.P.
B. a G.P.
C. an H.P.
D. none of these

Explanation

To solve the equation \( (1 + i\sqrt{3})^x - 2^x = 0 \) and determine the nature of its roots, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ (1 + i\sqrt{3})^x - 2^x = 0 \]
This can be rewritten as:
\[ (1 + i\sqrt{3})^x = 2^x \]

### Step 2: Divide both sides
Next, we divide both sides by \( 2^x \):
\[ \frac{(1 + i\sqrt{3})^x}{2^x} = 1 \]
This simplifies to:
\[ \left( \frac{1 + i\sqrt{3}}{2} \right)^x = 1 \]

### Step 3: Identify the complex number
Now, we need to express \( \frac{1 + i\sqrt{3}}{2} \) in polar form. The modulus \( r \) is given by:
\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]
The argument \( \theta \) can be found using:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Thus, we can write:
\[ \frac{1 + i\sqrt{3}}{2} = e^{i\frac{\pi}{3}} \]

### Step 4: Substitute back into the equation
Substituting this back, we have:
\[ \left(e^{i\frac{\pi}{3}}\right)^x = 1 \]
This implies:
\[ e^{i\frac{\pi}{3}x} = e^{i2k\pi} \quad \text{for } k \in \mathbb{Z} \]

### Step 5: Equate the exponents
From the equation above, we equate the exponents:
\[ \frac{\pi}{3}x = 2k\pi \]
Solving for \( x \):
\[ x = \frac{2k\pi}{\frac{\pi}{3}} = 6k \]

### Step 6: Determine the nature of the roots
The roots of the equation are given by:
\[ x = 6k \quad \text{for } k \in \mathbb{Z} \]
This means the roots are multiples of 6.

### Conclusion
The roots \( x = 6k \) form an arithmetic progression (AP) with a common difference of 6.