Question

write the expressions as a complex number in standard form.
-2(3+2i)

Answer 1

Just distribute the -2 outsisde the parentheses.

Answer 2

-6-4i

Answer 3

-2(3+2i) = -2 * 3 + (-2) * 2i = -6 - 4i

Answer 4

(4+3i)(1-i)

Answer 5

Here you use FOIL.

Answer 6

(4+3i)(1-i) =
4 * 1 + 4 * (-i) + 3i * 1 + 3i * (-i)

Answer 7

Now do all the multiplications and simplify.

Answer 8

is it 4+10i

Answer 9

(4+3i)(1-i) =
4 * 1 + 4 * (-i) + 3i * 1 + 3i * (-i) =
4 - 4i + 3i - 3i^2 =
4 -i - (3)(-1) =
4 - i + 3 =
7 - i

Answer 10

hi

Answer 11

so if you have a (1-i)(1+i) do they cancel oout

Answer 12

You have to either multiply it out using FOIL, or remember the product of a sum and a difference where the middle terms cancel out.
Let's use FOIL:
(1-i)(1+i) =
1 * 1 + 1 * i - i * 1 - i * i =
1 + i - i - i^2 =
1 - (-1) =
2

Answer 13

If you notice that (1-i)(1+i) is the product of a sum and a difference, and you remember that the product of a sum and a difference gives the difference of two squares, (a + b)(a - b) = a^2 - b^2, then you can use that:
(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2

Answer 14

i am going to show you a problem but dont help me with i just want you to tell me if its right