1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 5, 2008 #1
    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. jcsd
  3. Nov 5, 2008 #2

    Mark44

    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.)
     
  4. Nov 5, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  5. Nov 5, 2008 #4
    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.:)
     
  6. Nov 5, 2008 #5

    Mark44

    Staff: Mentor

    Re: mulitiplucation

    I'm sure you meant "distributive."

    Mark
     
  7. Nov 5, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: mulitiplucation

    Oh, dear, blush, blush! I just shouldn't use big words!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find (7 - j4) (4 - j3) - complex numbers
Loading...