Multiplying complex numbers?

Question

[And put it in a+bi form]

1.(3+i)(4+i)
2.(1+3i)(2-1)

you can answer one or bother( thad be great) but can someone just explain it thorughly b/c i have a test 2 morrow and im majorly comfused. Thankyou SOOO much<33

WAITTTT i have the answers to the other question. IF your going to answer. the last answer i need is
3.(2-i)(4+2i)
Omg u guys are the best Thanks<33xxx

Answer 1

it's easy juz do the multiplying like (a+b)(c+d) which equals to (ac+ad+bc+bd)....

For complex number.. i^2 = -1

so for Question 1 >>> (3+i)(4+i)

= 12+3i+4i+i^2
= 12+7i+(-1) >>> since i^2 = -1
= 11+7i >>> in the form of a+bi

For the second question >>> (1+3i)(2-1)

= (1+3i) (1)
= 1+3i

i think u forgot the i at the last part.... let says the question is >>> (1+3i)(2-1i)

= (2 - i + 6i - 3i^2)
= (2 + 5i -3(-1))
= (2+5i+3)
= (5+5i)

hope this help u to understand complex number.... good luck on ur test!!

Answer 2

To get the first one, use the FOIL rule, remembering
that i² = -1.
So we get 12 + 3i + 4i -1
Now combine the real and imaginary parts to get
11 + 7i.

For the second one, 2-1 = 1, so the answer is 1+3i.

Answer 3

First, expand it as if was any other fair of factors, and combine like terms:
(3 + i)(4 + i) = 12 + 4i + 3i + i^2 = 12 + 7i + i^2

Now remember that i = sqrt(-1), so i^2 = -1, and we have
12 + 7i + (-1) = 11 + 7i.

Answer 4

(3+i)(4+i)=3x4+4i+3i+i^2 = 12+(4+3)i-1 = 11+7i.

Remember that i^2 = -1 by definition.