Sure! Let's solve the complex multiplication step by step.
We need to compute the product of two complex numbers: [tex]$(6 + 4i)(3i)$[/tex].
1. First, we'll use the distributive property of multiplication over addition (also known as the FOIL method in algebra) to expand the product:
[tex]$(6 + 4i)(3i) = 6 \cdot 3i + 4i \cdot 3i.$[/tex]
2. Now, let's calculate each term individually:
- For the first term: [tex]\( 6 \cdot 3i \)[/tex]:
[tex]\[ 6 \cdot 3i = 18i. \][/tex]
- For the second term: [tex]\( 4i \cdot 3i \)[/tex]:
[tex]\[ 4i \cdot 3i = 12i^2. \][/tex]
3. Recall that [tex]\( i^2 = -1 \)[/tex]. Thus, we can simplify the second term:
[tex]\[ 12i^2 = 12(-1) = -12. \][/tex]
4. Next, we combine the simplified terms:
[tex]\[ 18i + (-12) = -12 + 18i. \][/tex]
5. Finally, we write the final expression in standard form for a complex number (a + bi).
The product of the complex numbers [tex]\( (6 + 4i) \)[/tex] and [tex]\( (3i) \)[/tex] is:
[tex]\[ -12 + 18i. \][/tex]
Comparing this result with the options given, the correct answer is:
[tex]\[ 12 + 18i. \][/tex]
