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Math Help - COmplex Analysis Problems

  1. #1
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    COmplex Analysis Problems

    Hi, having trouble with a few complex analysis problems. Here they are with my thoughts - I'd appreciate any help you can offer - thanks.

    1) Find all conformal mappings from the upper half plane to itself which fix the points at zero and infinity (easy to see this should be a linear transformation, but the constant is supposedly in the range (1, \infty ), which I'm not seeing.)

    2) Let f(x)=x^3+ax^2+bx+c with a,b,c real, a positive, and c negative. How many zeroes does $f(x)$ have in the left half plane? (Tried using Rouche but couldn't really get a conclusive answer.)

    3) Find a harmonic function on the intersection of a disk and the upper half plane (forgot the boundary conditions, etc. Just looking for a general map to a wedge-shaped region, the rest is elementary.)

    Thanks a ton.
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  2. #2
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    Re: COmplex Analysis Problems

    Quote Originally Posted by mstrfrdmx View Post
    2) Let f(x)=x^3+ax^2+bx+c with a,b,c real, a positive, and c negative. How many zeroes does $f(x)$ have in the left half plane? (Tried using Rouche but couldn't really get a conclusive answer.)
    No need for anything as advanced as Rouché. The sum of the roots is –a and is therefore negative. The product of the roots is –c and is therefore positive. If all three roots are real, those conditions imply that two of them are negative and one is positive. The only other possibility is that there is one real root and a pair of complex conjugate roots, say \alpha (real) and \beta \pm i\gamma (complex conjugate pair). Then the sum of the roots is \alpha+2\beta, and the product is \alpha(\beta^2+\gamma^2). Use that to work out how many of the roots have negative real part.
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  3. #3
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    Re: COmplex Analysis Problems

    For the first problem if we take f(z) =-\bar z, the f is a conformal mapping but it's not a linear transformation.
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    Re: COmplex Analysis Problems

    Quote Originally Posted by girdav View Post
    For the first problem if we take f(z) =-\bar z, the f is a conformal mapping but it's not a linear transformation.
    f(z)=-\overline{z} is nowhere holomorphic and hence cannot be conformal.
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    Super Member girdav's Avatar
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    Re: COmplex Analysis Problems

    Quote Originally Posted by hatsoff View Post
    f(z)=-\overline{z} is nowhere holomorphic and hence cannot be conformal.
    Yes, it's only an anti-holomorphic map. The definition I had of conformal map is a map which conserves the angle without taking care of the orientation, but the anti-holomorphic functions doesn't conserve the orientation of the angles.
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