The answer is C. 4/5 - 3/5 i
[tex] \frac{3-i}{3+i} = \frac{(3-i)(3-i)}{(3+i)(3-i)} \\ \\
(a+b)(a-b) = a^{2} -b^{2} \\ \\
\frac{(3-i)(3-i)}{(3+i)(3-i)}= \frac{ (3-i)^{2} }{3^{2}-i^{2} } \\ \\
(a-b)^{2} =a^{2}-2ab+b^{2} \\ \\
\frac{ (3-i)^{2} }{3^{2}-i^{2} } = \frac{3^{2}-2*3*i+i^{2} }{3^{2}-i^{2} } = \frac{9-6i+i^{2}}{9-i^{2}} \\ \\
i = \sqrt{-1} \\ i^{2}=-1 \\ \\
\frac{9-6i+i^{2}}{9-i^{2}} = \frac{9-6i-1}{9+1} = \frac{8-6i}{10}= \frac{8}{10}- \frac{6i}{10}= \frac{4}{5}- \frac{3}{5}i [/tex]
