Show me your expansion, and I will check it.
Did you remember that i^2 = -1 ?
Simplify the expression in a+bi form.
(3+i)(5-2i)(4-7i)
a. 67-145i
b. 75-123i
c. 53-145i
d. 61-123i
5 answers
I don't know how do to it. can you explain it to me?
if you can multiply binomials, you can multiply complex numbers. Start off with
(3+i)(5-2i)
= 3*5 - 3*2i + 5i - 2i^2
= 15 - 6i + 5i + 2 (since i^2 = -1)
= 17-i
Now multiply that by (4-7i) and you're done
(3+i)(5-2i)
= 3*5 - 3*2i + 5i - 2i^2
= 15 - 6i + 5i + 2 (since i^2 = -1)
= 17-i
Now multiply that by (4-7i) and you're done
Hi, I've just tried to work through this as you have and am really struggling. When I multiply by (4-7i) I get negative numbers?
maybe - what did you get?
take a look at
http://www.wolframalpha.com/input/?i=(3%2Bi)(5-2i)
and put in your numbers
take a look at
http://www.wolframalpha.com/input/?i=(3%2Bi)(5-2i)
and put in your numbers