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Math Help - Mobius transformation

  1. #1
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    Mobius transformation

    a) Find the mobius transformation which takes the points -1,1 and infinity to
    0,2 and 2i.

    I have done this bit, which gives me w=(2i+2ix)/(z-1+2i)

    b)desribe and find an equation for the real axis under the transformation found in part a.(i'm stuck on this and part d)


    c) Find the length of the path y(t) =i+3e^2it [0,pi/4]

    this gives me L=(pi/2)*root(2)

    d) now find an upper bound for the function g(z) =(z^2)e^z where y(t) is the path in part c.

    Any ideas or help would be appreciated. This is a past exam question and we have not been given solutions, which sucks.
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  2. #2
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    Quote Originally Posted by thespian View Post
    a) Find the mobius transformation which takes the points -1,1 and infinity to
    0,2 and 2i.

    I have done this bit, which gives me w=(2i+2ix)/(z-1+2i)

    b)desribe and find an equation for the real axis under the transformation found in part a.(i'm stuck on this and part d)

    Some ideas: we assume x \in \mathbb{R} and we try to find the image of w(x). Now, we assume a + bi \in Image(w(x)), and we find the restrictions for x:

    a+bi=\frac{2ix+2i}{x-1+2i}=\frac{4(x+1)+2(x^2-1)i}{(x-1)^2+4} \Longrightarrow  ax^2-2ax+5a+(bx^2-2bx+5b)i=4(x+1)+2(x^2-1)i

    Now compare real and imaginary parts in both sides and, surprisingly enough for me at least, we get that in order to get x's to solve the equations, it must be that \,1- \sqrt{2} \leq a\,,\,b \leq 1 + \sqrt{2}, so if I didn't make some mistake (and I wouldn't assume I didn't), the image of the real line under w is the above strip.




    c) Find the length of the path y(t) =i+3e^2it [0,pi/4]

    this gives me L=(pi/2)*root(2)

    d) now find an upper bound for the function g(z) =(z^2)e^z where y(t) is the path in part c.

    I don't get it: where in the equation g(z)=z^2e^z does y(t) enter? I can't see it.

    Tonio

    Any ideas or help would be appreciated. This is a past exam question and we have not been given solutions, which sucks.
    .
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  3. #3
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    Moius transform

    Thanks for the reply. Youre right part d doesnt make sense as it is, i will post the question properly as it is written.

    d)if g(z)=z^2*e^z and y(t) is the path in part c, find an upper bound for
    |∫g(z)dz|
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  4. #4
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    Quote Originally Posted by thespian View Post
    Thanks for the reply. Youre right part d doesnt make sense as it is, i will post the question properly as it is written.

    d)if g(z)=z^2*e^z and y(t) is the path in part c, find an upper bound for
    |∫g(z)dz|

    Now it's clear and we have an estimate:

    \left|\int\limits_{y(t)} g(z)dz\right| \leq Length(y(t))\cdot \max\limits_{[0,\frac{\pi}{4}]}|g(z)|

    Now remember that |g(z)|=|z|^2e^{Re(z)}

    Tonio
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  5. #5
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    Part c

    For part c we have:

    ax^2-2x(a+2)+5a-4=0

    using the formula b^2-4ac/2a i get:

    ax^2- x(2a+4)+ (5a-4)=0

    (2a+4)+-[(2a+4)^2-4a(5a-4)]^1/2]*(1/2a)

    (2a+4)=-[{24a-16a^2+16}^1/2]/2a

    and for "real " x i would need that the bit under the square root is>=0

    Namely:24a-16a^2+16>=0

    thus 16a^2-24a-16>=0

    a^2-(3/2)a-1>=0

    this has two solution for a:

    a= 2 and a=-1/2

    Thus to have real solutions for x that is only along the real axis we need -1/2<a<2

    I am trying to figure out how you got
    1-2^1/2<a<1+2^1/2

    can anyone else confirm which is correct?
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  6. #6
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    part c

    You were right, that is the right solution i just figured it out, thanks.
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