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Math Help - Need help with a question on complex number

  1. #1
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    Need help with a question on complex number

    It says to find 4 different complex number z1,z2,z3,z4 such that each complex number through these 4 properties will produce an integer.
    property 1) Re(z) = integer
    property 2)Im(z) = integer
    property 3) Re(z^(-1)) = integer
    property 4)Im(z^(-1)) = integer
    where Re and Im denotes the real and imaginary of the complex number respectively.

    Please help.
    I came up with only 1, i need 3 more.
    The one i came up with is
    z1=1+1i
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  2. #2
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    Quote Originally Posted by leonheart1234 View Post
    It says to find 4 different complex number z1,z2,z3,z4 such that each complex number through these 4 properties will produce an integer.
    property 1) Re(z) = integer
    property 2)Im(z) = integer
    property 3) Re(z^(-1)) = integer
    property 4)Im(z^(-1)) = integer
    where Re and Im denotes the real and imaginary of the complex number respectively.

    Please help.
    I came up with only 1, i need 3 more.
    The one i came up with is
    z1=1+1i


    This question was already solved some 1 week ago.

    Tonio
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  3. #3
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    Yes,I know its similar.But the notation is different.
    In the other post it says Re(z-1) and Im(z-1)
    Mine is Re(z^(-1)) and Im(z^(-1)).
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  4. #4
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    Quote Originally Posted by leonheart1234 View Post
    Yes,I know its similar.But the notation is different.
    In the other post it says Re(z-1) and Im(z-1)
    Mine is Re(z^(-1)) and Im(z^(-1)).

    Well, it's pretty similar, and if you did the past question and understood it then

    it should be clear what to do in this case.

    If z=x+iy\in\mathbb{C} , what is then z^{-1} ? After you answer this then try your problem.

    Tonio
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  5. #5
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    Quote Originally Posted by leonheart1234 View Post
    It says to find 4 different complex number z1,z2,z3,z4 such that each complex number through these 4 properties will produce an integer.
    property 1) Re(z) = integer property 2)Im(z) = integer
    property 3) Re(z^(-1)) = integer property 4)Im(z^(-1)) = integer
    Try: z_1=1,~z_2=i,~z_3=-1,~\&~z_4=-i~.

    Why do they work?
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  6. #6
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    let z = x + iy
    then z^{-1} = \frac{1}{z} = \frac{1}{x + iy}= \frac{1}{x + iy} * \frac{x - iy}{x - iy}= \frac{x - iy}{x^2 + y^2}= \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}
    then Re(z^{-1}) = \frac{x}{x^2 + y^2} ----------- (1)
    and
    Im(z^{-1}) = \frac{-y}{x^2 + y^2} --------------- (2)

    You need to find integers x and y such that (1) and (2) are also integers. Your solution of z1=1+1i with x = 1 and y = 1 is incorrect since (1) gives \frac{1}{2}.

    In fact you can go further and find all the possible solutions to your question.

    For x and y integers, x^2 + y^2 = k^2 \geq 1 must also be an integer. This equation is a circle centered about the origin and only has 4 solutions (along the x and y axes). x = k, y = 0 and x= -k, y=0 etc...

    for |x| > 1 or |y| > 1 , x^2 + y^2 >|x| or x^2 + y^2 > |y| so one of (1) or (2) will be a fraction. So
    |x| = 1 or |y|=1.

    HallsofIvy: either |x|= 1 and y= 0 or |y|= 1 and x= 0. Thus giving the four solutions Archie Moore gave.
    Last edited by Krahl; January 21st 2011 at 07:06 AM. Reason: Latex
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  7. #7
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    Quote Originally Posted by Krahl View Post
    let z = x + iy
    then z^{-1} = \frac{1}{z} = \frac{1}{x + iy}= \frac{1}{x + iy} * \frac{x - iy}{x - iy}= \frac{x - iy}{x^2 + y^2}= \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}
    then Re(z^{-1}) = \frac{x}{x^2 + y^2} ----------- (1)
    and
    Im(z^{-1}) = \frac{-y}{x^2 + y^2} --------------- (2)

    You need to find integers x and y such that (1) and (2) are also integers. Your solution of z1=1+1i with x = 1 and y = 1 is incorrect since (1) gives \frac{1}{2}.

    In fact you can go further and find all the possible solutions to your question.

    For x and y integers, x^2 + y^2 = k^2 \geq 1 must also be an integer. This equation is a circle centered about the origin and only has 4 solutions (along the x and y axes). x = k, y = 0 and x= -k, y=0 etc...

    for |x| > 1 or |y| > 1 , x^2 + y^2 >|x| or x^2 + y^2 > |y| so one of (1) or (2) will be a fraction. So
    |x| = 1 or |y|=1.
    You should say a bit more: either |x|= 1 and y= 0 or |y|= 1 and x= 0. Thus giving the four solutions Archie Moore gave.
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