Answer :
To solve the problem [tex]\((8 - 5i)^2\)[/tex], let's use the formula for squaring a binomial: [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex].
Here, [tex]\(a = 8\)[/tex] and [tex]\(b = 5i\)[/tex].
Let's plug these values into the formula:
1. Square the real part, [tex]\(a^2\)[/tex]:
[tex]\[8^2 = 64\][/tex]
2. Compute the cross term, [tex]\(-2ab\)[/tex]:
[tex]\[ -2 \cdot 8 \cdot 5i = -80i \][/tex]
3. Square the imaginary part, [tex]\(b^2\)[/tex]:
[tex]\[ (5i)^2 = 25i^2 \][/tex]
Since [tex]\(i^2 = -1\)[/tex], we have:
[tex]\[ 25i^2 = 25(-1) = -25 \][/tex]
Finally, add these results together:
[tex]\[ (8 - 5i)^2 = 64 - 80i - 25 \][/tex]
[tex]\[ = (64 - 25) - 80i \][/tex]
[tex]\[ = 39 - 80i \][/tex]
So, the simplified product is:
[tex]\[ \boxed{39 - 80i} \][/tex]
Here, [tex]\(a = 8\)[/tex] and [tex]\(b = 5i\)[/tex].
Let's plug these values into the formula:
1. Square the real part, [tex]\(a^2\)[/tex]:
[tex]\[8^2 = 64\][/tex]
2. Compute the cross term, [tex]\(-2ab\)[/tex]:
[tex]\[ -2 \cdot 8 \cdot 5i = -80i \][/tex]
3. Square the imaginary part, [tex]\(b^2\)[/tex]:
[tex]\[ (5i)^2 = 25i^2 \][/tex]
Since [tex]\(i^2 = -1\)[/tex], we have:
[tex]\[ 25i^2 = 25(-1) = -25 \][/tex]
Finally, add these results together:
[tex]\[ (8 - 5i)^2 = 64 - 80i - 25 \][/tex]
[tex]\[ = (64 - 25) - 80i \][/tex]
[tex]\[ = 39 - 80i \][/tex]
So, the simplified product is:
[tex]\[ \boxed{39 - 80i} \][/tex]
