Answer:
[tex]\large\boxed{1+i}[/tex]
Step-by-step explanation:
[tex]$\frac{4+2i }{3-i} $[/tex]
Factor the numerator
[tex]$\frac{2(2+i) }{3-i} $[/tex]
Multiply both numerator and denominator by the conjugate of the denominator.
[tex]$\frac{2(2+i) }{3-i} \cdot \frac{3+i }{3+i} = \frac{2(2+i)(3+i) }{9-i^2} $[/tex]
Once [tex]i^2=-1[/tex]
[tex]$\frac{2(2+i)(3+i) }{10} $[/tex]
Solving the numerator
[tex]2(2+i)(3+i) = 2(6+2i+3i+i^2) = 2(6+5i-1)=2(5+5i)[/tex]
[tex]$\frac{2(5+5i)}{2 \cdot 5} =\frac{5+5i}{5}=\boxed{1+i} $[/tex]
