Perform the operation and write the result in standard form.
(√19 + √20i)(√19 – √20i)
19 - 20i
@OOF I tried that. It's saying that it's wrong
i reworked it a bunch of times and i'm stumped. i'm sorry
since (a+b)(a-b) = a^2-b^2
(√19 + √20i)(√19 – √20i)
= 19 - 20i^2 = 19+20 = 39
(a-bi)(a+bi) = a^2+b^2
To perform the operation, we will use the exponentiation rule for complex numbers: (a + bi)(a - bi) = a^2 - b^2i^2.
First, let's simplify (√19 + √20i)(√19 – √20i):
= (√19)^2 - (√20i)^2
= 19 - 20i^2
Since i^2 = -1, we can substitute it into the equation:
= 19 - 20(-1)
= 19 + 20
= 39
So, the result is 39.