Complex Numbers Problem
Given \( x = \frac{7 - i}{2 - i} \) and \( y = \frac{13 - i}{4 + i} \), prove that \( x \) and \( y \) are complex conjugates, then show that \( x^2 + y^2 = 16 \).
Solution
\( x = \frac{7 - i}{2 - i} = \frac{(7 - i)(2 + i)}{(2 - i)(2 + i)} = \frac{14 + 7i - 2i - i^2}{4 - i^2} = \frac{14 + 5i + 1}{5} \)
\( = \frac{15 + 5i}{5} = 3 + i \)
\( y = \frac{13 - i}{4 + i} = \frac{(13 - i)(4 - i)}{(4 + i)(4 - i)} = \frac{52 - 13i - 4i + i^2}{16 - i^2} = \frac{52 - 17i - 1}{17} \)
\( = \frac{51 - 17i}{17} = 3 - i \)
\( \because x = 3 + i \) and \( y = 3 - i \), \( \therefore x \) and \( y \) are complex conjugates (same real part, imaginary parts have opposite signs).
\( x^2 = (3 + i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i \)
\( y^2 = (3 - i)^2 = 9 - 6i + i^2 = 9 - 6i - 1 = 8 - 6i \)
\( \therefore x^2 + y^2 = (8 + 6i) + (8 - 6i) = 16 \)