Certainly! Let's walk through the process of multiplying the complex numbers [tex]\((6 + \sqrt{-64})(3 - \sqrt{-16})\)[/tex] step-by-step.
First, we need to convert the square roots of the negative numbers into their complex forms. Recall that [tex]\(\sqrt{-1} = i\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit.
1. Calculate [tex]\(\sqrt{-64}\)[/tex]:
[tex]\[
\sqrt{-64} = \sqrt{64} \cdot \sqrt{-1} = 8i
\][/tex]
Thus, [tex]\(6 + \sqrt{-64} = 6 + 8i\)[/tex].
2. Calculate [tex]\(\sqrt{-16}\)[/tex]:
[tex]\[
\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i
\][/tex]
Thus, [tex]\(3 - \sqrt{-16} = 3 - 4i\)[/tex].
Now we need to multiply the two complex numbers [tex]\( (6 + 8i) \)[/tex] and [tex]\( (3 - 4i) \)[/tex]:
3. Use the distributive property (also known as the FOIL method for binomials):
[tex]\[
(6 + 8i)(3 - 4i) = 6 \cdot 3 + 6 \cdot (-4i) + 8i \cdot 3 + 8i \cdot (-4i)
\][/tex]
4. Calculate each term:
[tex]\[
6 \cdot 3 = 18
\][/tex]
[tex]\[
6 \cdot (-4i) = -24i
\][/tex]
[tex]\[
8i \cdot 3 = 24i
\][/tex]
[tex]\[
8i \cdot (-4i) = -32i^2
\][/tex]
5. Combine the results from step 4:
[tex]\[
18 - 24i + 24i - 32i^2
\][/tex]
6. Notice that the imaginary parts [tex]\(-24i\)[/tex] and [tex]\(+24i\)[/tex] cancel each other out:
[tex]\[
18 - 32i^2
\][/tex]
7. Recall that [tex]\(i^2 = -1\)[/tex]. Therefore, [tex]\(-32i^2 = -32(-1) = 32\)[/tex]:
[tex]\[
18 + 32
\][/tex]
8. Finally, add the real parts:
[tex]\[
18 + 32 = 50
\][/tex]
So, the product of [tex]\((6 + \sqrt{-64})(3 - \sqrt{-16})\)[/tex] is:
[tex]\[
\boxed{50}
\][/tex]
