Question

If \( \alpha \) denotes the number of solutions of \( |1 - i|^x = 2^x \) and \( \beta = \frac{|z|}{\arg(z)} \), where \[ z = \frac{\pi}{4} (1 + i)^4 \left( \frac{1 - \sqrt{\pi} i}{\sqrt{\pi} + i} + \frac{\sqrt{\pi} - i}{1 + \sqrt{\pi} i} \right), \quad i = \sqrt{-1}, \] then the distance of the point \( (\alpha, \beta) \) from the line \( 4x - 3y = 7 \) is _____

Answer 1

Correct Answer: 3

Given:
The following series of calculations are provided. Let's go step by step: 

Step 1:
\((\sqrt{2})^x = 2^x \Rightarrow x = 0 \Rightarrow \alpha = 1\)

Step 2:
We have \( z = \frac{\pi}{4} (1 + i)^4 \left[ \frac{\sqrt{\pi - \pi i - \sqrt{\pi}}}{\pi + 1} + \frac{\sqrt{\pi - i - \pi i - \sqrt{\pi}}}{1 + \pi} \right] \), which simplifies to \( z = -\frac{\pi}{2} (1 + 4i + 6i^2 + 4i^3 + 1) = 2\pi \).

Step 3:
Next, we calculate \( \beta = \frac{2\pi}{\pi} = 4 \).

Step 4:
For the distance between point (1, 4) and the line \(4x - 3y = 7\), we use the distance formula:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Substituting \( A = 4, B = -3, C = -7 \) and point (x₁, y₁) = (1, 4), we calculate:

\[ d = \frac{|4(1) - 3(4) - 7|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 - 12 - 7|}{5} = \frac{15}{5} = 3 \]

Final Answer: The distance from the point (1, 4) to the line \(4x - 3y = 7\) is \( 3 \).

Answer 2

\[ (\sqrt{2})^x = 2^x \implies x = 0 \implies \alpha = 1 \]

\[ z = \frac{\pi}{4}(1 + i)^4 \]
 \[ = \frac{-\pi i}{2} \left[ \sqrt{\frac{\pi - \pi i - \sqrt{\pi}}{\pi + 1}} + \sqrt{\frac{\pi - i - \pi i - \sqrt{\pi}}{1 + \pi}} \right] \] 
\[ = \frac{-\pi i}{2} \left( 1 + 4i + 6i^2 + 4i^3 + 1 \right) \] 
\[ = 2\pi i \beta = \frac{2\pi}{\pi/2} = 4 \]

Distance from \( (1, 4) \) to \( 4x - 3y = 7 \) will be:

\[ \frac{15}{5} = 3 \]