The number of non-zero integral sol ution of the e quation ` |1- 2i ^(x)=5^(x)` is

Options

A. Zero (No solution)
B. One
C. Two
D. Three

Explanation

To solve the equation \( |1 - 2i^x| = 5^x \) for the number of non-zero integral solutions, we will follow these steps:

### Step 1: Understand the equation
We start with the equation:
\[ |1 - 2i^x| = 5^x \]

### Step 2: Calculate the modulus
The modulus of a complex number \( a + bi \) is given by:
\[ |a + bi| = \sqrt{a^2 + b^2} \]
In our case, we have \( a = 1 \) and \( b = -2i^x \). Therefore, we need to find \( |1 - 2i^x| \).

### Step 3: Express \( i^x \)
Recall that \( i = e^{i\frac{\pi}{2}} \). Thus, we can express \( i^x \) as:
\[ i^x = e^{i\frac{\pi}{2}x} \]
This means that \( |i^x| = 1 \) for any integer \( x \).

### Step 4: Calculate \( |1 - 2i^x| \)
Since \( |i^x| = 1 \), we have:
\[ |1 - 2i^x| = |1 - 2 \cdot e^{i\frac{\pi}{2}x}| \]
To find this modulus, we can express it as:
\[ |1 - 2 \cdot (\cos(\frac{\pi}{2}x) + i \sin(\frac{\pi}{2}x)| = |1 - 2 \cos(\frac{\pi}{2}x) + 2i \sin(\frac{\pi}{2}x)| \]
Calculating the modulus:
\[ = \sqrt{(1 - 2 \cos(\frac{\pi}{2}x))^2 + (2 \sin(\frac{\pi}{2}x))^2} \]

### Step 5: Simplify the expression
Expanding the modulus:
\[ = \sqrt{(1 - 2 \cos(\frac{\pi}{2}x))^2 + 4 \sin^2(\frac{\pi}{2}x)} \]
Using the identity \( \sin^2 + \cos^2 = 1 \):
\[ = \sqrt{1 - 4 \cos(\frac{\pi}{2}x) + 4 \cos^2(\frac{\pi}{2}x) + 4 \sin^2(\frac{\pi}{2}x)} \]
\[ = \sqrt{1 - 4 \cos(\frac{\pi}{2}x) + 4} \]
\[ = \sqrt{5 - 4 \cos(\frac{\pi}{2}x)} \]

### Step 6: Set up the equation
Now we equate this to \( 5^x \):
\[ \sqrt{5 - 4 \cos(\frac{\pi}{2}x)} = 5^x \]

### Step 7: Square both sides
Squaring both sides gives:
\[ 5 - 4 \cos(\frac{\pi}{2}x) = 25^x \]
Rearranging gives:
\[ 4 \cos(\frac{\pi}{2}x) = 5 - 25^x \]

### Step 8: Analyze the equation
The left side, \( 4 \cos(\frac{\pi}{2}x) \), varies between -4 and 4, while the right side, \( 5 - 25^x \), decreases as \( x \) increases.

### Step 9: Find integral solutions
We need to find when \( 5 - 25^x \) is within the range of -4 to 4.

1. For \( x = 1 \):
\[ 5 - 25^1 = 5 - 25 = -20 \quad (\text{not valid}) \]
2. For \( x = 0 \):
\[ 5 - 25^0 = 5 - 1 = 4 \quad (\text{valid}) \]
3. For \( x = -1 \):
\[ 5 - 25^{-1} = 5 - \frac{1}{25} = \frac{124}{25} \quad (\text{valid}) \]

Continuing this process shows that there are no non-zero integral solutions.

### Conclusion
Thus, the number of non-zero integral solutions to the equation \( |1 - 2i^x| = 5^x \) is:
\[ \boxed{0} \]