# Find (7 - j4) (4 - j3) - complex numbers

1. Nov 5, 2008

### morbello

hello im working on a book on maths and i have come to a part i need help with.

(7-j4)(4-j3) is the equation i know that a^2-2ab+b^2 but what is the formula for abcd

the answer i have to get to is 28-j37-12 from what i know allready i got the 28 and the 12

so if you can help or explain it to me.

2. Nov 5, 2008

### Staff: Mentor

Re: mulitiplucation

(7-j4)(4-j3) isn't an equation; it's an expression to be multiplied. When you multiply two binomials (expressions with two terms) you're going to get four products. For this product:
(a + b)(c + d)
you get a*c + a*d + b*c + b*d

It works the same with your problem, but in addition you should simplify any resulting expressions with j^2, which is equal to -1. (Mathematicians usually call the imaginary unit i, sqrt(-1), while engineers usually call it j.)

3. Nov 5, 2008

### HallsofIvy

Staff Emeritus
Re: mulitiplucation

You mean, I take it, that you want to multiply (7- 4j)(4- 3j). Writing the numbers after the variables is confusing- some people will write "j4" to mean "j4". You, of course, wrote "a^2" so you know better, but still...

You also write "i know that a^2-2ab+b^2" which is simply nonsense! I assume you mean that you know "(a- b)(a- b)= a^2- 2ab+ b^2". But do you know how to get that? That's the crucial point. Also, you do NOT "want to get to 28-j37-12", you have mis-copied. What you want to get to is 28- 37j- 12j^2.

"Multiplication is associative"- that simply means that a(b+ c)= ab+ bc which is true for numbers or algebraic expressions. 2(3+ 5)= 2(8)= 16, of course, but also 2(3+ 5)= 2(3)+ 2(5)= 6+ 10= 16.

(a- b)(a- b)= a(a-b)- b(a-b) is one application of that. Using it again, a(a-b)= a^2- ab and b(a-b)= ab- b^2. Putting those together, (a- b)(a- b)= (a^2- ab)-(ab- b^2)= a^2- 2ab+ b^2.

Now try that with (7- 4j)(4- 3j). That is equal to 7(4- 3j)- 4j(4- 3j). 7(4- 3j)= 7(4)- 7(3j)= 28- 21j. 4j(4- 3j)= 4j(4)- 4j(3j)= 16j- 13j^2. Putting those together, (7- 4j)(4- 3j)= (28- 21j)- (16j- 12j^2)= 28- (21-16)j+ 12j^2= 28- 37j+ 12j^2.

4. Nov 5, 2008

### morbello

Re: mulitiplucation

thank you it has helped im sure if i try the exersies in the book now. I will be able to do them.:)

5. Nov 5, 2008

### Staff: Mentor

Re: mulitiplucation

I'm sure you meant "distributive."

Mark

6. Nov 5, 2008

### HallsofIvy

Staff Emeritus
Re: mulitiplucation

Oh, dear, blush, blush! I just shouldn't use big words!