The solution of the equation ` (1 + i sqrt(3))^(x) = 2^(x)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none

Answer

To solve the equation \( (1 + i \sqrt{3})^x = 2^x \), we will follow these steps:

### Step 1: Convert the complex number to polar form
The complex number \( 1 + i \sqrt{3} \) can be expressed in polar form. We first find its modulus and argument.

**Modulus**:
\[ r = |1 + i \sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

**Argument**:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]

Thus, we can write:
\[ 1 + i \sqrt{3} = 2 \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} \right) \]

### Step 2: Rewrite the equation using polar form
Now substituting the polar form into the equation gives:
\[ (2 (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}))^x = 2^x \]

This simplifies to:
\[ 2^x \left( \cos\left(\frac{\pi}{3} x\right) + i \sin\left(\frac{\pi}{3} x\right) \right) = 2^x \]

### Step 3: Divide both sides by \( 2^x \)
Assuming \( 2^x \neq 0 \), we can divide both sides by \( 2^x \):
\[ \cos\left(\frac{\pi}{3} x\right) + i \sin\left(\frac{\pi}{3} x\right) = 1 \]

### Step 4: Set the real and imaginary parts
For the equality to hold, the imaginary part must be zero, and the real part must equal one:
1. \( \cos\left(\frac{\pi}{3} x\right) = 1 \)
2. \( \sin\left(\frac{\pi}{3} x\right) = 0 \)

### Step 5: Solve for \( x \)
From \( \cos\left(\frac{\pi}{3} x\right) = 1 \):
\[ \frac{\pi}{3} x = 2k\pi \quad \text{for } k \in \mathbb{Z} \]
This simplifies to:
\[ x = 6k \]

From \( \sin\left(\frac{\pi}{3} x\right) = 0 \):
\[ \frac{\pi}{3} x = n\pi \quad \text{for } n \in \mathbb{Z} \]
This simplifies to:
\[ x = 3n \]

### Step 6: Combine results
Since \( x = 6k \) and \( x = 3n \) must hold simultaneously, we can express \( n \) in terms of \( k \):
\[ 3n = 6k \implies n = 2k \]

Thus, the solutions can be expressed as:
\[ x = 6k \quad \text{where } k \in \mathbb{Z} \]

### Conclusion
The solutions of the equation \( (1 + i \sqrt{3})^x = 2^x \) are of the form \( x = 6k \), where \( k \) is any integer.

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