which is the product of the complex numbers (2 + i) and (3 - 2i)?\na. 6 - 2i\nb. 4 - i\nc. 8 - i\nd. 8…
which is the product of the complex numbers (2 + i) and (3 - 2i)?\na. 6 - 2i\nb. 4 - i\nc. 8 - i\nd. 8 - 7i\ne. 4 - 7i
Answer
Explanation:
Step1: Use FOIL method
$(2 + i)(3 - 2i)=2\times3+2\times(-2i)+i\times3 + i\times(-2i)$
Step2: Simplify each term
$2\times3 = 6$, $2\times(-2i)=-4i$, $i\times3 = 3i$, $i\times(-2i)=-2i^{2}$. Since $i^{2}=-1$, then $-2i^{2}=-2\times(-1) = 2$.
Step3: Combine like - terms
$6-4i + 3i+2=(6 + 2)+(-4i+3i)=8 - i$
Answer:
C. $8 - i$