Given that `(1 + i)/(1 + 2^(2)i) xx (1 + 3^(2) i)/(1 + 4^(2) i) xx….xx (1 + (2n-1)^(2)i)/(1+(2n)^(2)i)= (a + bi)/(c+di)`, show that `(2)/(17) xx (82)/(257) xx ….xx ((2n-1)^(4) + 1)/((2n)^(4) + 1) = (a^(2) + b^(2))/(c^(2) + d^(2))`

Answer

To solve the problem, we need to show that

\[ \frac{2}{17} \times \frac{82}{257} \times \ldots \times \frac{(2n-1)^4 + 1}{(2n)^4 + 1} = \frac{a^2 + b^2}{c^2 + d^2} \]

where

\[ \frac{(1 + i)}{(1 + 2^2 i)} \times \frac{(1 + 3^2 i)}{(1 + 4^2 i)} \times \ldots \times \frac{(1 + (2n-1)^2 i)}{(1 + (2n)^2 i)} = \frac{a + bi}{c + di} \]

### Step-by-Step Solution:

1. **Express the given product in terms of magnitudes**:
We start with the expression:

\[ \frac{(1 + i)}{(1 + 2^2 i)} \times \frac{(1 + 3^2 i)}{(1 + 4^2 i)} \times \ldots \times \frac{(1 + (2n-1)^2 i)}{(1 + (2n)^2 i)} \]

Taking the magnitudes of both sides, we have:

\[ \left| \frac{(1 + i)}{(1 + 2^2 i)} \times \frac{(1 + 3^2 i)}{(1 + 4^2 i)} \times \ldots \times \frac{(1 + (2n-1)^2 i)}{(1 + (2n)^2 i)} \right| = \frac{|a + bi|}{|c + di|} \]

2. **Calculate the magnitudes**:
The magnitude of a complex number \( z = x + yi \) is given by \( |z| = \sqrt{x^2 + y^2} \).

- For \( 1 + i \):
\[ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

- For \( 1 + 2^2 i = 1 + 4i \):
\[ |1 + 4i| = \sqrt{1^2 + 4^2} = \sqrt{17} \]

- For \( 1 + 3^2 i = 1 + 9i \):
\[ |1 + 9i| = \sqrt{1^2 + 9^2} = \sqrt{82} \]

- For \( 1 + 4^2 i = 1 + 16i \):
\[ |1 + 16i| = \sqrt{1^2 + 16^2} = \sqrt{257} \]

Continuing this pattern, we find the magnitudes for all terms up to \( n \).

3. **Combine the magnitudes**:
Thus, we can express the product of the magnitudes as:

\[ \frac{|1 + i|}{|1 + 4i|} \times \frac{|1 + 9i|}{|1 + 16i|} \times \ldots = \frac{\sqrt{2}}{\sqrt{17}} \times \frac{\sqrt{82}}{\sqrt{257}} \times \ldots \]

This leads to:

\[ \frac{\sqrt{2} \times \sqrt{82} \times \ldots}{\sqrt{17} \times \sqrt{257} \times \ldots} \]

4. **Square both sides**:
Squaring both sides to eliminate the square roots gives:

\[ \frac{2 \times 82 \times \ldots}{17 \times 257 \times \ldots} = \frac{a^2 + b^2}{c^2 + d^2} \]

5. **Final expression**:
Therefore, we conclude that:

\[ \frac{2}{17} \times \frac{82}{257} \times \ldots \times \frac{(2n-1)^4 + 1}{(2n)^4 + 1} = \frac{a^2 + b^2}{c^2 + d^2} \]