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Math Help - Adding and Subtracting Complex Numbers

  1. #1
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    Adding and Subtracting Complex Numbers

    Ok. I understand that a complex number can be written in the form...a+bi.
    Which leads me to the idea that
    (5-6i)-4i is equivalent to (5-6i)-(0-4i).
    Which gives me 5-2i as a solution, because 5-0=5 and -6i-(-4i) is -2i. So..
    (5-6i)-4i = 5-2i
    Is this a true statement?
    My text says no, and I can't wrap my head around it.

    P.S. (a+bi)-(c+di)= a+bi-c-di = (a-c)+(b-d)i
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  2. #2
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    Quote Originally Posted by EyesForEars View Post
    Ok. I understand that a complex number can be written in the form...a+bi.
    Which leads me to the idea that
    (5-6i)-4i is equivalent to (5-6i)-(0-4i).

    The above should be written (5-6i)-4i = (5-6i) - (0 + 4i). Remember, when you put grouping symbols around an expression and precede the group with a negative sign, each sign within the group must be changed to its opposite.

    -4i = -(+4i)



    Which gives me 5-2i as a solution, because 5-0=5 and -6i-(-4i) is -2i. So.. (5-6i)-4i = 5-2i

    Removing the parentheses on the above we get 5 - 6i - 4i = 5 - 10i

    Is this a true statement?
    My text says no, and I can't wrap my head around it.

    P.S. (a+bi)-(c+di)= a+bi-c-di = (a-c)+(b-d)i

    Using this, we get (5-6i)-(0+4i) = 5-6i-0-4i = (5-0)+(-6-4)i = 5-10i
    ..
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  3. #3
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    So, any time I apply grouping symbols to an expression preceeded by a negative sign, I have to then reverse the signs of said terms. Got it.

    I think I got it.

    So, I'll say.

    (19-26i)-7i = (19-26i)-(0+7i) = 19-26i-0-7i = 19-33i
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  4. #4
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    Quote Originally Posted by EyesForEars View Post
    So, any time I apply grouping symbols to an expression preceeded by a negative sign, I have to then reverse the signs of said terms. Got it.

    I think I got it.

    So, I'll say.

    (19-26i)-7i = (19-26i)-(0+7i) = 19-26i-0-7i = 19-33i
    You are correct, my friend!!
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  5. #5
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    dont forget "--" = +

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