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Math Help - Elementary transformations (FP3)

  1. #1
    Junior Member rednest's Avatar
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    Post Elementary transformations (FP3)

    Points P and Q represent the complex numbers w and z respectively in the Argand diagram.

    If w=\frac{1+zi}{z+i}, z \neq -i and w=u+iv, z=x+iy, express u and v in terms of x and y.

    Prove that when P describes the portion of the imaginary axis between the points representing -i and i, Q describes the whole of the positive imaginary axis.
    I don't get what the question means. Please help me understand it.
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  2. #2
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    Quote Originally Posted by rednest View Post
    I don't get what the question means. Please help me understand it.
    Unfortunately for you, the first part means that you have

    u + iv = \frac{1 - y + ix}{x + iy + i}

    and you have to

    (i) equate the real part of the left hand side with the real part of the right hand side

    (ii) equate the imaginary part of the left hand side with the imaginary part of the right hand side.
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  3. #3
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    Quote Originally Posted by rednest View Post
    I don't get what the question means. Please help me understand it.
    I hope there is a simple way out. I had to do it by solving a strange system of equations.

    So before writing out the entire solution, I want to check if my answer is right.

    Is
    u = \frac{xy}{x+x^2 + (1+y)^2}

    v = \frac{x+x^2 + 1+y}{x+x^2 + (1+y)^2}??


    EDIT: So TES proved its wrong
    Last edited by Isomorphism; June 19th 2008 at 09:15 PM.
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  4. #4
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    Quote Originally Posted by rednest View Post
    I don't get what the question means. Please help me understand it.

    u+iv=\frac{1+(x+iy)i}{x+iy+i}=\frac{1-y+ix}{x+(y+1)i}

    Now multiplying by the conjugate on the left side we get

    u+iv=\frac{1-y+ix}{x+(y+1)i}\cdot \left( \frac{x-(y+1)i}{x-(y+1)i} \right)=\frac{x(1-y)-(1-y)(y+1)i+x^2i-x(y+1)}{x^2-(y+1)^2}

    u+iv=\frac{-2xy-(1+x^2-y^2)i}{x-(y+1)^2}=\frac{-2xy}{x-(y+1)^2}+\frac{-(1+x^2-y^2)}{x-(y+1)^2}i

    equating the real and immaginary parts we get

    u(x,y)=\frac{-2xy}{x-(y+1)^2} \\\ \\\ v(x,y)=\frac{-(1+x^2-y^2)}{x-(y+1)^2}

    We are concerned with the part of the plane -1 \le y \le 1 \\\ x=0 (The immaginary axis between [-i,i])

    u(0,y)=\frac{-2(0)y}{0-(y+1)^2}=0 \\\ \\\ v(0,y)=\frac{-(1+0^2-y^2)}{0-(y+1)^2}=\frac{-(1-y)(1+y)}{-(y+1)^2}=\frac{1-y}{1+y}

    So v(0,y) for  y \in [-1,1] maps to

    [0,\infty)

    Yeah!! Thanks isomorphism!!
    Last edited by TheEmptySet; June 19th 2008 at 09:16 PM.
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    Quote Originally Posted by TheEmptySet View Post
    u+iv=\frac{1-y+ix}{x+(y+1)i}\cdot \left( \frac{x-(y+1)i}{x-(y+1)i} \right)=\color{red}{\frac{x(1-y)+(1-y)(y+1)i+x^2i-x(y+1)}{x^2-(y+1)^2}}
    It should be \frac{x(1-y)-(1-y)(y+1)i+x^2i+x(y+1)}{x^2+(y+1)^2}
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    Quote Originally Posted by Isomorphism View Post
    It should be \frac{x(1-y)-(1-y)(y+1)i+x^2i+x(y+1)}{x^2+(y+1)^2}
    Awsome this should fix my mapping

    I will fix my above post
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    u+iv=\frac{1+(x+iy)i}{x+iy+i}=\frac{1-y+ix}{x+(y+1)i}

    Now multiplying by the conjugate on the left side we get

    u+iv=\frac{1-y+ix}{x+(y+1)i}\cdot \left( \frac{x-(y+1)i}{x-(y+1)i} \right)=\color{red}{\frac{x(1-y)-(1-y)(y+1)i+x^2i-x(y+1)}{x^2-(y+1)^2}}
    Hmm shouldnt it be

    u+iv = \frac{x(1-y)-(1-y)(y+1)i+x^2i+x(y+1)}{x^2+(y+1)^2}??

    --------------------------------------------------------------------------------------------------

    Here is the solution anyway:

    Thus u = \frac{2x}{x^2+(y+1)^2} and v =\frac{x^2 + y^2 - 1}{x^2+(y+1)^2}.

    Now P between i and -i means 1 < v < -1 and u = 0.

    This means x = 0 and \left|\frac{x^2 + y^2 - 1}{x^2+(y+1)^2}\right| < 1


    \left|\frac{x^2 + y^2 - 1}{x^2+(y+1)^2}\right| < 1 \Rightarrow |x^2 + y^2 - 1|< |x^2+(y+1)^2|

    Now since x = 0,

    |y^2 - 1| < |(y+1)^2|
    |y+1| > 0, so divide both sides by it

    |y - 1| < |y+1|

    Its easy to prove that the above inequality holds only for y > 0.

    But this means x=0, |v| < 1 \Rightarrow x= 0,y = \text{Im}(Q) > 0 and thus when " describes the portion of the imaginary axis between the points representing and , describes the whole of the positive imaginary axis."
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    We can write g(z)=\frac{1+zi}{z+i} = \frac{1 + (z+i) - i}{z+i}.
    How is \frac{1+zi}{z+i} = \frac{1 + (z+i) - i}{z+i}?
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  9. #9
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    Quote Originally Posted by Isomorphism View Post
    How is \frac{1+zi}{z+i} = \frac{1 + (z+i) - i}{z+i}?
    When you had 4 shots of vodka. Sorry .
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  10. #10
    Junior Member rednest's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    When you had 4 shots of vodka. Sorry .
    Lol. Nice
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