Respuesta :
The question is given below as
[tex](3+4i)(3-i)[/tex]Concept:
Apply the complex arithmetic rule
[tex]\begin{gathered} (a+bi)(c+di) \\ =(ac-bd)+(ad+bc)i \end{gathered}[/tex]By comparing coefficient with the main question, we will have
[tex]a=3,b=4,c=3,d=-1[/tex]Step 1: Substitute the values in the arithmetic rule
[tex]\begin{gathered} (a+bi)(c+di) \\ (ac-bd)+(ad+bc)i \\ (3\times3)-(4\times-1)+(3\times-1)+(4\times3)i \\ =(9+4)+(-3+12)i \\ =13+9i \end{gathered}[/tex]Alternatively,
Use the FOIL method
[tex]\begin{gathered} (3+4i)(3-i) \\ 3(3-i)+4i(3-i) \\ =9-3i+12i-4i^2 \\ =9+9i-4i^2 \\ \text{note :} \\ i^2=-1 \\ =9+9i-4(-1) \\ =9+9i+4 \\ =9+4+9i \\ =13+9i \end{gathered}[/tex]Hence,
The final answer is = 13 + 9i
