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Math Help - Complex number help??

  1. #1
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    Complex number help??

    Find "x" and "Y" if

    1) 2(x+yi) = x-yi

    2) (X+2i)(1-i) = 5+yi

    Thanks, please also do the process of getting the answer not just the answer.
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  2. #2
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    Quote Originally Posted by jumko View Post
    Find "x" and "Y" if

    1) 2(x+yi) = x-yi

    Mr F says: Expand and equate the real and imaginary parts on each side:

    2x + 2yi = x - yi. Therefore

    Equate real parts: 2x = x .... (1)

    Equate imaginary parts: 2y = -y .... (2)

    Therefore x = y = 0.



    2) (X+2i)(1-i) = 5+yi

    Mr F says: Use the same approach as above.

    Thanks, please also do the process of getting the answer not just the answer.
    ..
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  3. #3
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    Quote Originally Posted by jumko View Post
    1) 2(x+yi) = x-yi
    2x + 2yi = x - yi

    (2x - x) + (2y + y)i = 0

    For this to be true for arbitrary x and y then we must have that the real part is exactly 0 and the imaginary part is exactly 0.

    So
    2x - x = 0
    and
    2y + y = 0

    The solution here is trivial, but the method shown here is the way to do all of these.

    -Dan
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  4. #4
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    thanks.

    But, if it wouldn't be any trouble could you show the process of another slightly more harder one?

    (x+yi)(2+i) = 2x-(y+1)i
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  5. #5
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    Quote Originally Posted by jumko View Post
    thanks.

    But, if it wouldn't be any trouble could you show the process of another slightly more harder one?

    (x+yi)(2+i) = 2x-(y+1)i
    Well, how about you do a few things for me - I'll help you where required:

    1. Expand the left hand side:

    (x + yi)(2 + i) = ...........

    2. After expanding the left hand side, state:
    (a) the real part of the left hand side.
    (b) the imaginary part of the left hand side.

    3. Set up the following equations:

    real part of left hand side = 2x .... (1)
    imaginary part of left hand side = -(y + 1) = -y - 1 .... (2)

    4. Solve equations (1) and (2) above simultaneously for x and y.
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