Option A is correct.
The given complex number is
[tex]\frac{3+i}{3-i}[/tex]
Multiply the numerator and denominator by 3+i and solve as follows:
[tex]\begin{gathered} \frac{3+i}{3-1}\times\frac{3+i}{3-i}=\frac{(3+i)^2}{3^2-i^2} \\ =\frac{3^2+i^2+6i}{9+1} \\ =\frac{9-1+6i}{10} \\ =\frac{8+6i}{10} \\ =\frac{8}{10}+\frac{6}{10}i \\ =\frac{4}{5}+\frac{3}{5}i \end{gathered}[/tex]
