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Math Help - Equation, Complex numbers

  1. #1
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    Equation, Complex numbers

    Hello.

    Is given:

    $(1+i)^2-(2-\text{i})^2=a+bi\;\;\;\text{find:  a+b}

    $A)\;1

    $B)\;2

    $C)\;3

    $D)\;4


    I have tried:

    $(1+i)^2-(2-i)^2=a+bi

    $1^2+2*1*\text{i}+\text{i}^2-(2^2-2*2*\text{i}+\text{i}^2)=a+bi

    $2i-4+4i+1=a+bi

    6i-3=a+bi

    3(2i-1)=a+bi

    Here I'm stuck... Any help would be great. Thanks
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  2. #2
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    Re: Equation, Complex numbers

    Quote Originally Posted by kreshnik View Post
    $(1+i)^2-(2-\text{i})^2=a+bi\;\;\;\text{find:  a+b}
    $A)\;1
    $B)\;2
    $C)\;3
    $D)\;4
    Here is a hint:
    If a+bi=-1+2i then a=-1~\&~b=2.
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  3. #3
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    Re: Equation, Complex numbers

    I never wrote this:
    Quote Originally Posted by plato View Post
    here is a hint:
    If a+bi=-1+2i then a=-1~\&~b=2.
    you meant:

    Quote Originally Posted by kreshnik View Post
    3(2i-1)=a+bi
    or
    Quote Originally Posted by kreshnik View Post
    6i-3=a+bi
    so: a+bi=-3+6i

    hence: a=-3\;\;\;,\;\;\;b=6

    Am I wrong?

    a+b=-3+6=3
    or
    a+b=-1+2=1

    which one?, sorry for taking so long
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Equation, Complex numbers

    Quote Originally Posted by kreshnik

    a+b=-3+6=3
    This is correct. The're no two options! Where does your other 'solution' come from?
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  5. #5
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    Re: Equation, Complex numbers

    Quote Originally Posted by Siron View Post
    This is correct. The're no two options! Where does your other 'solution' come from?
    I think Plato somehow didn't see number 3. But it's ok, now I know how to do It, thanks everyone.
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