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Math Help - Problem with complexe numbers.

  1. #1
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    Problem with complexe numbers.

    Hi !
    I am have a problem resolving an equation with complexe numbers on their algebrique form.

    Here is it :

    For all complexe z = x + iy (x and y real and different from -i) we have :
    Z = (z - 1) / (iz + 1)
    Find Re(Z) and Im(Z) by x and y.

    I already have the answer : Re(Z) = (x+y-1)/((1-y)^2 + x) and Im(Z) = (-x^2-y^2+x+y)/((1-y)^2+x^2, but I don't know how to get to it.

    I've tried to resolv it with but it doesn't seems to work :
    Z = (z-1)/(iz+1)
    = (x+iy-1)/(i(x+iy)+1)
    = (x+iy-1)/(ix-y-1) , because i^2 = -1
    = (x-iy-1)/(i(x-iy)+1)
    = (x-iy-1)/(ix+y-1)

    And Re(Z)=0 <=> Z=
    So : Z- = 0
    (x+iy-1)/(ix-y-1) - (x-iy-1)/(ix+y-1) = 0
    ((x+iy-1)(ix+y-1)-(x-iy-1)(ix-y-1))/((ix-y-1)(ix+y-1)) = 0
    (ix^2+xy-x-xy+iy^2-iy-ix-y+1)-(ix^2-xy-x+xy+iy^2+iy-ix-y+1)/((ix-y-1)(ix+y-1))=0
    but : (ix^2-ix^2+iy^2-iy^2-ix+ix-iy-iy+xy-xy+xy-xy-x+x-y+y+1-1) = 0

    I still don't know what is Re(Z) I must have done a mistake and I don't know where or how to find the answer.

    Help me understand this problem! Please.

    Thanks.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by meghan View Post
    Hi !
    I am have a problem resolving an equation with complexe numbers on their algebrique form.

    Here is it :

    For all complexe z = x + iy (x and y real and different from -i) we have :
    Z = (z - 1) / (iz + 1)
    Find Re(Z) and Im(Z) by x and y.

    I already have the answer : Re(Z) = (x+y-1)/((1-y)^2 + x) and Im(Z) = (-x^2-y^2+x+y)/((1-y)^2+x^2, but I don't know how to get to it.

    I've tried to resolv it with but it doesn't seems to work :
    Z = (z-1)/(iz+1)
    = (x+iy-1)/(i(x+iy)+1)
    = (x+iy-1)/(ix-y-1) , because i^2 = -1
    = (x-iy-1)/(i(x-iy)+1) no, because the denominator contains another \boldface{i}
    = (x-iy-1)/(ix+y-1)

    And Re(Z)=0 <=> Z=
    So : Z- = 0
    (x+iy-1)/(ix-y-1) - (x-iy-1)/(ix+y-1) = 0
    ((x+iy-1)(ix+y-1)-(x-iy-1)(ix-y-1))/((ix-y-1)(ix+y-1)) = 0
    (ix^2+xy-x-xy+iy^2-iy-ix-y+1)-(ix^2-xy-x+xy+iy^2+iy-ix-y+1)/((ix-y-1)(ix+y-1))=0
    but : (ix^2-ix^2+iy^2-iy^2-ix+ix-iy-iy+xy-xy+xy-xy-x+x-y+y+1-1) = 0

    I still don't know what is Re(Z) I must have done a mistake and I don't know where or how to find the answer.

    Help me understand this problem! Please.

    Thanks.
    Z=\displaystyle\frac{(x-1)+iy}{i(x+iy)+1}=\frac{(x-1)+iy}{ix+i^2y+1}=\frac{(x-1)+iy}{(1-y)+ix}

    Now you could use the complex conjugate of the denominator to make it real, so multiply by 1=\displaystyle\frac{(1-y)-ix}{(1-y)-ix}

    \displaystyle\frac{\left[(x-1)+iy\right]\left[(1-y)-ix\right]}{\left[(1-y)+ix\right]\left[(1-y)-ix\right]}

    Then continue...
    Last edited by Archie Meade; November 8th 2010 at 03:56 AM. Reason: awful typo
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  3. #3
    MHF Contributor
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    sorry, there was a nasty typo in my post....edited.
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