-1 or 0 when Multiplying complex numbers?

Discussion in 'Math' started by Biggsy100, Jun 10, 2015.

  1. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    So I have applied the FOIL method to multiply my intial question which is:

    If A = 1 + j2 and B = 2 - j3

    (a+bj) (c+dj) = ac +adj + bcj +bdi^2

    so,

    (1 +j2)+(c+dj) = (1 x 2) + (1 x 3) + (2 x 2) + (2 x 3)

    = 2 + 4j + 4j +6j^2 = This is where my problem comes.......

    =3 + 4j +4j + -6 ? I am assuming the - is to satisfy B equation?

    Going through this method I reach an answer with single digits....5

    I would assume A second number to show real and imaginary parts. Is this correct or am I missing something?
     
    Last edited: Jun 10, 2015
  2. tjohnson

    Active Member

    Dec 23, 2014
    614
    121
    That's not correct. If the FOIL method is applied correctly to multiply your initial question, you should obtain the answer 8 + j1.

    You are correct that (a+bj)(c+dj) = ac + adj + bcj + bdj^2. This shows that you understand the FOIL method, but it should be easier to solve individual problems by using FOIL on them rather than by using the formula you derived.

    So applying the FOIL method to A = 1 + j2 and B = 2 - j3:
    (1+j2)(2-j3) = (1\times 2) + (1\times -j3) + (j2\times 2) + (j2\times -j3) = 2 - 3j + 4j - 6j^2 = (2+6) + j(-3+4) = 8 + j1
     
    Last edited: Jun 10, 2015
  3. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,791
    829
    Be careful in your steps. For example, in one place you have used both i and j to represent the imaginary terms.

    Secondly, where did 4j+4j come from and where did they go?
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,418
    488
    Hi,

    I had given a more complete explanation of multiplication in the other thread which you might want to check out first.

    When you deal with complex numbers they each have a real and imaginary part, and you must always be aware of what you are doing with each part and keep it separated at least until the last step.

    For multiplication, you have to do four multiplications, and for each one you must determine whether it is real or imaginary, and if it is imaginary (even if temporarily) you must use some form of the operator 'j'. You never ignore that.

    So for the general expression:
    (a+b*j)*(c+d*j)

    we get:
    a*c, which is real,
    a*d*j, which is imaginary,
    b*j*c, which is imaginary, and
    b*j*d*j, which is a power of j because they both have a 'j', and this leads to b*d*j^2.

    So the four terms added together are:
    a*c+a*d*j+b*j*c+b*d*j^2

    and if you look in the other thread for the table of powers of j you'll see:
    j^1=j
    j^2=-1
    j^3=-j
    j^4=1

    and since we have j^2 in our result so far we can convert that into:
    a*c+a*d*j+b*j*c+b*d*(-1)

    and all we did so far was we replaced that j^2 with a minus 1 (-1).
    Now since that made that term real also, we can add that to the other real term and we get:
    a*c-b*d
    for the real part, and:
    (a*d+b*c)
    for the imaginary part, so combined we have:
    a*c-b*d+(a*d+b*c)*j

    and note we put the 'j' back when we wrote it out as all one number.
    This brings up the other little point about what the imaginary part really is.
    For a number like:
    a+b*j

    the imaginary part is 'b' alone, not 'b*j', the 'j' acts as a placeholder to show that 'b' is the imaginary part. The real part is of course 'a'.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Look at this and ask yourself if it makes any sense at all.

    First, you have no multiplication at all on the left hand side.
    Second, where do the 3s come from?
    Third, where do the 'c' and 'd' terms go?
    Fourth, where do the j's go?

    Then in your next line you add (1 x 3) to get 4 instead of multiplying them to get 3.

    You continue to be incredibly sloppy in your work. That is going to continue to hamper your efforts to learn this stuff until you decide to stop that.

    Now, I'm guessing, from the right hand side, that you meant the left hand side to be

    (1+j2)*(2+j3)

    You asked if the - sign is there to satisfy B equation. What is "B equation"? We are NOT mind readers! Stop making us guess at what you are doing or referring to.

    The - sign is there because j² = -1 BY DEFINITION!

    You then proceed to combine 3 + 4j +4j + -6 to get 5, which means that you simply ignored the fact that two of the terms are multiplied by j. It has been stated over and over and over in your various threads that you can't do that. Is (3+4x) equal to 7? If not, then why would you think that (3+j4) is?

    You aren't ready to deal with complex numbers yet because your algebra skills are extremely weak. Math builds on itself and you are building on a poor foundation. You need to take a course (a formal course, an online tutorial, or study on your own) in basic algebra otherwise you will just be digging yourself a deeper and deeper hole.
     
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